Chapter 11 Conic Sections

Given:

The centre of the circle is at the origin, i.e., $(h, k) = (0, 0)$.

The radius of the circle is $r$.

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To Find:

An equation of the circle.

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Solution:

The general equation of a circle with centre at $(h, k)$ and radius $r$ is given by:

$(x - h)^2 + (y - k)^2 = r^2$

We are given that the centre of the circle is at the origin, which means $h = 0$ and $k = 0$.

Substitute these values into the general equation of the circle:

$(x - 0)^2 + (y - 0)^2 = r^2$

Simplifying the equation, we get:

This is the required equation of the circle with the centre at $(0, 0)$ and radius $r$.

Given:

The centre of the circle is $(h, k) = (-3, 2)$.

The radius of the circle is $r = 4$.

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To Find:

The equation of the circle.

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Solution:

The standard equation of a circle with centre $(h, k)$ and radius $r$ is given by:

$(x - h)^2 + (y - k)^2 = r^2$

Substitute the given values of $h = -3$, $k = 2$, and $r = 4$ into the equation:

$(x - (-3))^2 + (y - 2)^2 = 4^2$

Simplify the expression:

$(x + 3)^2 + (y - 2)^2 = 16$

This is the required equation of the circle.

Given:

The equation of the circle is $x^2 + y^2 + 8x + 10y – 8 = 0$.

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To Find:

The centre and the radius of the circle.

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Solution:

The standard equation of a circle is $(x - h)^2 + (y - k)^2 = r^2$, where $(h, k)$ is the centre and $r$ is the radius.

We will convert the given equation into this standard form by completing the square.

Given equation:

$x^2 + y^2 + 8x + 10y – 8 = 0$

First, group the x-terms and y-terms together and move the constant to the right side:

$(x^2 + 8x) + (y^2 + 10y) = 8$

To complete the square for the x-terms, take half of the coefficient of $x$ (which is 8), square it, and add it to both sides. Half of 8 is 4, and $4^2 = 16$.

To complete the square for the y-terms, take half of the coefficient of $y$ (which is 10), square it, and add it to both sides. Half of 10 is 5, and $5^2 = 25$.

$(x^2 + 8x + 16) + (y^2 + 10y + 25) = 8 + 16 + 25$

Now, we can write the expressions in the parentheses as perfect squares:

$(x + 4)^2 + (y + 5)^2 = 49$

Now, we compare this with the standard form $(x - h)^2 + (y - k)^2 = r^2$.

By comparing $(x - (-4))^2 + (y - (-5))^2 = 7^2$, we get:

$h = -4$

$k = -5$

$r^2 = 49 \implies r = \sqrt{49} = 7$

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Therefore, the centre of the circle is $(-4, -5)$ and the radius is 7.

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Alternate Solution

The general equation of a circle is $x^2 + y^2 + 2gx + 2fy + c = 0$.

The centre is given by $(-g, -f)$ and the radius is given by $r = \sqrt{g^2 + f^2 - c}$.

Comparing the given equation $x^2 + y^2 + 8x + 10y – 8 = 0$ with the general form:

$2g = 8 \implies g = 4$

$2f = 10 \implies f = 5$

$c = -8$

Now, we can find the centre and radius:

Centre $= (-g, -f) = (-4, -5)$

Radius $r = \sqrt{g^2 + f^2 - c} = \sqrt{(4)^2 + (5)^2 - (-8)}$

$r = \sqrt{16 + 25 + 8} = \sqrt{49} = 7$

The results from both methods are the same.

Given:

A circle passes through two points, A(2, –2) and B(3, 4).

The centre of the circle, let's call it C(h, k), lies on the line $x + y = 2$.

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To Find:

The equation of the circle.

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Solution:

Let the centre of the circle be C(h, k).

Since the centre lies on the line $x + y = 2$, its coordinates must satisfy the line's equation.

The distance from the centre of a circle to any point on its circumference is equal to the radius. Therefore, the distance CA must be equal to the distance CB.

$CA = CB \implies CA^2 = CB^2$

Using the distance formula, $(x_2 - x_1)^2 + (y_2 - y_1)^2$, we can write:

$CA^2 = (h - 2)^2 + (k - (-2))^2 = (h - 2)^2 + (k + 2)^2$

$CB^2 = (h - 3)^2 + (k - 4)^2$

Now, we set $CA^2 = CB^2$:

$(h - 2)^2 + (k + 2)^2 = (h - 3)^2 + (k - 4)^2$

Expand both sides:

$(h^2 - 4h + 4) + (k^2 + 4k + 4) = (h^2 - 6h + 9) + (k^2 - 8k + 16)$

Cancel $h^2$ and $k^2$ from both sides:

$-4h + 4k + 8 = -6h - 8k + 25$

Now, bring all the h and k terms to one side and constants to the other:

$-4h + 6h + 4k + 8k = 25 - 8$

We now have a system of two linear equations (i) and (ii) to solve for h and k.

From equation (i), we have $h = 2 - k$.

Substitute this into equation (ii):

$2(2 - k) + 12k = 17$

$4 - 2k + 12k = 17$

$10k = 13 \implies k = \frac{13}{10}$

Now find h:

$h = 2 - k = 2 - \frac{13}{10} = \frac{20 - 13}{10} = \frac{7}{10}$

So, the centre of the circle is C$(\frac{7}{10}, \frac{13}{10})$.

Next, we find the radius squared, $r^2$, by calculating the distance from the centre C to one of the points, say A(2, -2).

$r^2 = CA^2 = (h - 2)^2 + (k + 2)^2$

$r^2 = (\frac{7}{10} - 2)^2 + (\frac{13}{10} + 2)^2$

$r^2 = (\frac{7 - 20}{10})^2 + (\frac{13 + 20}{10})^2$

$r^2 = (\frac{-13}{10})^2 + (\frac{33}{10})^2$

$r^2 = \frac{169}{100} + \frac{1089}{100} = \frac{1258}{100} = \frac{629}{50}$

The standard equation of the circle is $(x - h)^2 + (y - k)^2 = r^2$.

Substituting the values of h, k, and $r^2$:

$\left(x - \frac{7}{10}\right)^2 + \left(y - \frac{13}{10}\right)^2 = \frac{629}{50}$

To write this in general form, we expand the equation:

$x^2 - \frac{7}{5}x + \frac{49}{100} + y^2 - \frac{13}{5}y + \frac{169}{100} = \frac{629}{50}$

$x^2 + y^2 - \frac{7}{5}x - \frac{13}{5}y + \frac{218}{100} = \frac{1258}{100}$

$x^2 + y^2 - \frac{7}{5}x - \frac{13}{5}y + \frac{218 - 1258}{100} = 0$

$x^2 + y^2 - \frac{7}{5}x - \frac{13}{5}y - \frac{1040}{100} = 0$

$x^2 + y^2 - \frac{7}{5}x - \frac{13}{5}y - \frac{52}{5} = 0$

Multiplying by 5 to clear the denominators, we get:

$5x^2 + 5y^2 - 7x - 13y - 52 = 0$

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The equation of the circle is $5x^2 + 5y^2 - 7x - 13y - 52 = 0$.

Given:

The centre of the circle is at $(h, k) = (0, 2)$.

The radius of the circle is $r = 2$.

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To Find:

The equation of the circle.

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Solution:

The standard equation of a circle with centre $(h, k)$ and radius $r$ is given by:

$(x - h)^2 + (y - k)^2 = r^2$

Substitute the given values $h = 0$, $k = 2$, and $r = 2$ into the equation:

$(x - 0)^2 + (y - 2)^2 = 2^2$

Simplify the equation:

$x^2 + (y - 2)^2 = 4$

Expanding the term $(y - 2)^2$:

$x^2 + (y^2 - 4y + 4) = 4$

Rearrange the terms to get the general form:

$x^2 + y^2 - 4y + 4 - 4 = 0$

This is the required equation of the circle.

Given:

The centre of the circle is at $(h, k) = (-2, 3)$.

The radius of the circle is $r = 4$.

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To Find:

The equation of the circle.

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Solution:

The standard equation of a circle with centre $(h, k)$ and radius $r$ is given by:

$(x - h)^2 + (y - k)^2 = r^2$

Substitute the given values $h = -2$, $k = 3$, and $r = 4$ into the equation:

$(x - (-2))^2 + (y - 3)^2 = 4^2$

Simplify the equation:

This is the required equation of the circle in standard form.

Alternatively, we can expand this to the general form:

$(x + 2)^2 = x^2 + 4x + 4$

$(y - 3)^2 = y^2 - 6y + 9$

Substitute these expansions back into equation (i):

$x^2 + 4x + 4 + y^2 - 6y + 9 = 16$

Rearrange the terms:

$x^2 + y^2 + 4x - 6y + 4 + 9 - 16 = 0$

$x^2 + y^2 + 4x - 6y + 13 - 16 = 0$

This is the equation of the circle in general form.

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The equation of the circle is $(x + 2)^2 + (y - 3)^2 = 16$ or $x^2 + y^2 + 4x - 6y - 3 = 0$.

Given:

The centre of the circle is at $(h, k) = \left(\frac{1}{2}, \frac{1}{4}\right)$.

The radius of the circle is $r = \frac{1}{12}$.

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To Find:

The equation of the circle.

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Solution:

The standard equation of a circle with centre $(h, k)$ and radius $r$ is given by:

$(x - h)^2 + (y - k)^2 = r^2$

Substitute the given values $h = \frac{1}{2}$, $k = \frac{1}{4}$, and $r = \frac{1}{12}$ into the equation:

$\displaystyle \left(x - \frac{1}{2}\right)^2 + \left(y - \frac{1}{4}\right)^2 = \left(\frac{1}{12}\right)^2$

Simplify the equation:

This is the required equation of the circle in standard form.

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Alternate Solution (General Form):

We can expand the standard form equation (i) to the general form $Ax^2 + By^2 + Cx + Dy + E = 0$.

Expand the squared terms:

$\displaystyle \left(x - \frac{1}{2}\right)^2 = x^2 - 2x\left(\frac{1}{2}\right) + \left(\frac{1}{2}\right)^2 = x^2 - x + \frac{1}{4}$

$\displaystyle \left(y - \frac{1}{4}\right)^2 = y^2 - 2y\left(\frac{1}{4}\right) + \left(\frac{1}{4}\right)^2 = y^2 - \frac{1}{2}y + \frac{1}{16}$

Substitute these expansions back into equation (i):

$\displaystyle x^2 - x + \frac{1}{4} + y^2 - \frac{1}{2}y + \frac{1}{16} = \frac{1}{144}$

Group terms and move the constant term to the left side:

$\displaystyle x^2 + y^2 - x - \frac{1}{2}y + \frac{1}{4} + \frac{1}{16} - \frac{1}{144} = 0$

Find a common denominator for the constant terms, which is 144.

$\displaystyle \frac{1}{4} = \frac{1 \times 36}{4 \times 36} = \frac{36}{144}$

$\displaystyle \frac{1}{16} = \frac{1 \times 9}{16 \times 9} = \frac{9}{144}$

$\displaystyle \frac{1}{144} = \frac{1}{144}$

Combine the constant terms:

$\displaystyle \frac{36}{144} + \frac{9}{144} - \frac{1}{144} = \frac{36 + 9 - 1}{144} = \frac{45 - 1}{144} = \frac{44}{144}$

Simplify the fraction $\frac{44}{144}$ by dividing both numerator and denominator by their greatest common divisor, which is 4.

$\displaystyle \frac{44}{144} = \frac{\cancel{44}^{11}}{\cancel{144}_{36}} = \frac{11}{36}$

So the equation becomes:

$\displaystyle x^2 + y^2 - x - \frac{1}{2}y + \frac{11}{36} = 0$

Multiply the entire equation by 36 to eliminate denominators:

$\displaystyle 36\left(x^2 + y^2 - x - \frac{1}{2}y + \frac{11}{36}\right) = 36(0)$

$\displaystyle 36x^2 + 36y^2 - 36x - 36\left(\frac{1}{2}\right)y + 36\left(\frac{11}{36}\right) = 0$

$\displaystyle 36x^2 + 36y^2 - 36x - \cancel{36}^{18}\left(\frac{1}{\cancel{2}_{1}}\right)y + 11 = 0$

This is the equation of the circle in general form.

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The equation of the circle can be written in standard form as $\left(x - \frac{1}{2}\right)^2 + \left(y - \frac{1}{4}\right)^2 = \frac{1}{144}$ or in general form as $36x^2 + 36y^2 - 36x - 18y + 11 = 0$.

Given:

The centre of the circle is at $(h, k) = (1, 1)$.

The radius of the circle is $r = \sqrt{2}$.

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To Find:

The equation of the circle.

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Solution:

The standard equation of a circle with centre $(h, k)$ and radius $r$ is given by:

$(x - h)^2 + (y - k)^2 = r^2$

Substitute the given values $h = 1$, $k = 1$, and $r = \sqrt{2}$ into the equation:

$(x - 1)^2 + (y - 1)^2 = (\sqrt{2})^2$

Simplify the equation:

This is the required equation of the circle in standard form.

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Alternate Solution (General Form):

We can expand the standard form equation (i) to the general form $Ax^2 + By^2 + Cx + Dy + E = 0$.

Expand the squared terms:

$(x - 1)^2 = x^2 - 2x(1) + 1^2 = x^2 - 2x + 1$

$(y - 1)^2 = y^2 - 2y(1) + 1^2 = y^2 - 2y + 1$

Substitute these expansions back into equation (i):

$x^2 - 2x + 1 + y^2 - 2y + 1 = 2$

Rearrange the terms and move the constant to the left side:

$x^2 + y^2 - 2x - 2y + 1 + 1 - 2 = 0$

$x^2 + y^2 - 2x - 2y + 2 - 2 = 0$

This is the equation of the circle in general form.

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The equation of the circle can be written in standard form as $(x - 1)^2 + (y - 1)^2 = 2$ or in general form as $x^2 + y^2 - 2x - 2y = 0$.

Given:

The centre of the circle is at $(h, k) = (-a, -b)$.

The radius of the circle is $r = \sqrt{a^2 - b^2}$.

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To Find:

The equation of the circle.

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Solution:

The standard equation of a circle with centre $(h, k)$ and radius $r$ is given by:

$(x - h)^2 + (y - k)^2 = r^2$

Substitute the given values $h = -a$, $k = -b$, and $r = \sqrt{a^2 - b^2}$ into the equation:

$(x - (-a))^2 + (y - (-b))^2 = (\sqrt{a^2 - b^2})^2$

Simplify the equation:

This is the required equation of the circle in standard form.

Note that for the radius to be real, we must have $a^2 - b^2 \ge 0$, or $a^2 \ge b^2$.

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Alternate Solution (General Form):

We can expand the standard form equation (i) to the general form $Ax^2 + By^2 + Cx + Dy + E = 0$.

Expand the squared terms:

$(x + a)^2 = x^2 + 2ax + a^2$

$(y + b)^2 = y^2 + 2by + b^2$

Substitute these expansions back into equation (i):

$x^2 + 2ax + a^2 + y^2 + 2by + b^2 = a^2 - b^2$

Rearrange the terms and move the constant term from the right side to the left side:

$x^2 + y^2 + 2ax + 2by + a^2 + b^2 - (a^2 - b^2) = 0$

$x^2 + y^2 + 2ax + 2by + a^2 + b^2 - a^2 + b^2 = 0$

Combine like terms ($a^2$ cancels out):

This is the equation of the circle in general form.

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The equation of the circle can be written in standard form as $(x + a)^2 + (y + b)^2 = a^2 - b^2$ or in general form as $x^2 + y^2 + 2ax + 2by + 2b^2 = 0$, provided that $a^2 \ge b^2$.

Given:

The equation of the circle is $(x + 5)^2 + (y – 3)^2 = 36$.

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To Find:

The centre and the radius of the given circle.

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Solution:

The given equation of the circle is:

The standard equation of a circle with centre at $(h, k)$ and radius $r$ is given by:

Comparing equation (i) with the standard equation (ii), we can write equation (i) as:

$(x - (-5))^2 + (y - 3)^2 = 6^2$

By comparing the corresponding terms, we can find the values of $h$, $k$, and $r^2$:

$x - h$ corresponds to $x - (-5)$, so $h = -5$.

$y - k$ corresponds to $y - 3$, so $k = 3$.

$r^2$ corresponds to $36$.

$r^2 = 36$

To find the radius $r$, take the square root of $r^2$. Since the radius must be a non-negative value, we take the positive root:

$r = \sqrt{36}$

$r = 6$

Therefore, the centre of the circle is $(h, k) = (-5, 3)$ and the radius is $r = 6$.

Given:

The equation of the circle is $x^2 + y^2 – 4x – 8y – 45 = 0$.

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To Find:

The centre and the radius of the given circle.

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Solution:

The given equation of the circle is:

$x^2 + y^2 – 4x – 8y – 45 = 0$

To find the centre and radius, we can rewrite this equation in the standard form $(x - h)^2 + (y - k)^2 = r^2$ by completing the square for the terms involving $x$ and $y$.

Group the terms with $x$ and $y$, and move the constant term to the right side of the equation:

$(x^2 - 4x) + (y^2 - 8y) = 45$

To complete the square for the x-terms, take half of the coefficient of $x$ (which is -4), square it $( (-4/2)^2 = (-2)^2 = 4 )$, and add it inside the parenthesis and to the right side.

To complete the square for the y-terms, take half of the coefficient of $y$ (which is -8), square it $( (-8/2)^2 = (-4)^2 = 16 )$, and add it inside the parenthesis and to the right side.

$(x^2 - 4x + 4) + (y^2 - 8y + 16) = 45 + 4 + 16$

Now, rewrite the expressions in the parentheses as perfect squares:

$(x - 2)^2 + (y - 4)^2 = 65$

This equation is now in the standard form $(x - h)^2 + (y - k)^2 = r^2$.

Comparing $(x - 2)^2 + (y - 4)^2 = 65$ with the standard form $(x - h)^2 + (y - k)^2 = r^2$, we can identify $h$, $k$, and $r^2$:

$x - h$ corresponds to $x - 2$, so $h = 2$.

$y - k$ corresponds to $y - 4$, so $k = 4$.

$r^2$ corresponds to $65$.

$r^2 = 65$

To find the radius $r$, take the square root of $r^2$. Since the radius must be a non-negative value, we take the positive root:

$r = \sqrt{65}$

Therefore, the centre of the circle is $(h, k) = (2, 4)$ and the radius is $r = \sqrt{65}$.

Given:

The equation of the circle is $x^2 + y^2 – 8x + 10y – 12 = 0$.

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To Find:

The centre and the radius of the given circle.

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Solution:

The given equation of the circle is:

$x^2 + y^2 – 8x + 10y – 12 = 0$

To find the centre and radius, we can rewrite this equation in the standard form $(x - h)^2 + (y - k)^2 = r^2$ by completing the square for the terms involving $x$ and $y$.

Group the terms with $x$ and $y$, and move the constant term to the right side of the equation:

$(x^2 - 8x) + (y^2 + 10y) = 12$

To complete the square for the x-terms, take half of the coefficient of $x$ (which is -8), square it $( (-8/2)^2 = (-4)^2 = 16 )$, and add it inside the parenthesis and to the right side.

To complete the square for the y-terms, take half of the coefficient of $y$ (which is 10), square it $( (10/2)^2 = 5^2 = 25 )$, and add it inside the parenthesis and to the right side.

$(x^2 - 8x + 16) + (y^2 + 10y + 25) = 12 + 16 + 25$

Now, rewrite the expressions in the parentheses as perfect squares:

$(x - 4)^2 + (y + 5)^2 = 53$

This equation is now in the standard form $(x - h)^2 + (y - k)^2 = r^2$.

Comparing $(x - 4)^2 + (y + 5)^2 = 53$ with the standard form $(x - h)^2 + (y - k)^2 = r^2$, we can write the equation as:

$(x - 4)^2 + (y - (-5))^2 = 53$

By comparing the corresponding terms, we can identify $h$, $k$, and $r^2$:

$x - h$ corresponds to $x - 4$, so $h = 4$.

$y - k$ corresponds to $y - (-5)$, so $k = -5$.

$r^2$ corresponds to $53$.

$r^2 = 53$

To find the radius $r$, take the square root of $r^2$. Since the radius must be a non-negative value, we take the positive root:

$r = \sqrt{53}$

Therefore, the centre of the circle is $(h, k) = (4, -5)$ and the radius is $r = \sqrt{53}$.

Given:

The equation of the circle is $2x^2 + 2y^2 – x = 0$.

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To Find:

The centre and the radius of the given circle.

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Solution:

The given equation of the circle is:

$2x^2 + 2y^2 – x = 0$

To find the centre and radius, we first need to rewrite this equation in the standard general form $x^2 + y^2 + 2Gx + 2Fy + C = 0$ or the standard form $(x - h)^2 + (y - k)^2 = r^2$.

Divide the entire equation by 2 to make the coefficients of $x^2$ and $y^2$ equal to 1:

$\displaystyle \frac{2x^2}{2} + \frac{2y^2}{2} - \frac{x}{2} = \frac{0}{2}$

$\displaystyle x^2 + y^2 - \frac{1}{2}x = 0$

Now, we complete the square for the terms involving $x$ and $y$. Group the terms:

$\displaystyle \left(x^2 - \frac{1}{2}x\right) + y^2 = 0$

To complete the square for the x-terms, take half of the coefficient of $x$ (which is $-\frac{1}{2}$), square it $\left(\left(\frac{-1/2}{2}\right)^2 = \left(-\frac{1}{4}\right)^2 = \frac{1}{16}\right)$, and add it inside the parenthesis and to the right side.

For the y-terms, since there is no linear $y$ term, $y^2$ is already a perfect square $(y - 0)^2$.

$\displaystyle \left(x^2 - \frac{1}{2}x + \frac{1}{16}\right) + (y^2 - 0)^2 = 0 + \frac{1}{16}$

Now, rewrite the expression in the first parenthesis as a perfect square and simplify the right side:

This equation is now in the standard form $(x - h)^2 + (y - k)^2 = r^2$.

Comparing equation (i) with the standard form $(x - h)^2 + (y - k)^2 = r^2$, we can identify $h$, $k$, and $r^2$:

$x - h$ corresponds to $x - \frac{1}{4}$, so $h = \frac{1}{4}$.

$y - k$ corresponds to $y - 0$, so $k = 0$.

$r^2$ corresponds to $\frac{1}{16}$.

$r^2 = \frac{1}{16}$

To find the radius $r$, take the square root of $r^2$. Since the radius must be a non-negative value, we take the positive root:

$\displaystyle r = \sqrt{\frac{1}{16}}$

$\displaystyle r = \frac{1}{4}$

Therefore, the centre of the circle is $(h, k) = \left(\frac{1}{4}, 0\right)$ and the radius is $r = \frac{1}{4}$.

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Alternate Solution (Using General Form Coefficients):

The given equation is $2x^2 + 2y^2 – x = 0$.

Divide by 2 to get the general form $x^2 + y^2 + 2Gx + 2Fy + C = 0$:

$\displaystyle x^2 + y^2 - \frac{1}{2}x = 0$

Compare this to $x^2 + y^2 + 2Gx + 2Fy + C = 0$:

$2G = -\frac{1}{2} \implies G = -\frac{1}{4}$

$2F = 0 \implies F = 0$

$C = 0$

The centre of the circle is $(-G, -F)$.

Centre $= \left(-\left(-\frac{1}{4}\right), -0\right) = \left(\frac{1}{4}, 0\right)$

The radius of the circle is $r = \sqrt{G^2 + F^2 - C}$.

$\displaystyle r = \sqrt{\left(-\frac{1}{4}\right)^2 + (0)^2 - 0}$

$\displaystyle r = \sqrt{\frac{1}{16} + 0 - 0}$

$\displaystyle r = \sqrt{\frac{1}{16}}$

$\displaystyle r = \frac{1}{4}$

Both methods yield the same results.

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The centre of the circle is $\left(\frac{1}{4}, 0\right)$ and the radius is $\frac{1}{4}$.

Given:

The circle passes through the points $P(4, 1)$ and $Q(6, 5)$.

The centre of the circle lies on the line $4x + y = 16$.

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To Find:

The equation of the circle.

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Solution:

Let the centre of the circle be $C(h, k)$ and the radius be $r$.

The standard equation of a circle is given by:

$(x - h)^2 + (y - k)^2 = r^2$

Since the centre $C(h, k)$ lies on the line $4x + y = 16$, it must satisfy the equation of the line:

Since the circle passes through the point $P(4, 1)$, the distance from the centre $C(h, k)$ to $P(4, 1)$ is equal to the radius $r$. Using the distance formula:

$r^2 = (4 - h)^2 + (1 - k)^2$

Expanding this equation:

$r^2 = (16 - 8h + h^2) + (1 - 2k + k^2)$

Since the circle also passes through the point $Q(6, 5)$, the distance from the centre $C(h, k)$ to $Q(6, 5)$ is also equal to the radius $r$:

$r^2 = (6 - h)^2 + (5 - k)^2$

Expanding this equation:

$r^2 = (36 - 12h + h^2) + (25 - 10k + k^2)$

Equating the expressions for $r^2$ from equations (2) and (3):

$h^2 + k^2 - 8h - 2k + 17 = h^2 + k^2 - 12h - 10k + 61$

Subtract $h^2 + k^2$ from both sides:

$-8h - 2k + 17 = -12h - 10k + 61$

Move all terms involving $h$ and $k$ to the left side and constant terms to the right side:

$-8h + 12h - 2k + 10k = 61 - 17$

$4h + 8k = 44$

Divide the entire equation by 4:

Now we have a system of two linear equations with two variables, $h$ and $k$, from equations (1) and (4):

1) $4h + k = 16$

4) $h + 2k = 11$

From equation (1), we can express $k$ in terms of $h$:

Substitute equation (5) into equation (4):

$h + 2(16 - 4h) = 11$

$h + 32 - 8h = 11$

Combine like terms:

$-7h + 32 = 11$

$-7h = 11 - 32$

$-7h = -21$

$\displaystyle h = \frac{-21}{-7}$

$h = 3$

Now substitute the value of $h$ back into equation (5) to find $k$:

$k = 16 - 4(3)$

$k = 16 - 12$

$k = 4$

So, the centre of the circle is $(h, k) = (3, 4)$.

Now, we need to find the radius squared, $r^2$. We can use equation (2) and substitute the values of $h = 3$ and $k = 4$:

$\displaystyle r^2 = (3)^2 + (4)^2 - 8(3) - 2(4) + 17$

$r^2 = 9 + 16 - 24 - 8 + 17$

$r^2 = 25 - 32 + 17$

$r^2 = -7 + 17$

$r^2 = 10$

Alternatively, using equation (3) with $h = 3$ and $k = 4$:

$\displaystyle r^2 = (3)^2 + (4)^2 - 12(3) - 10(4) + 61$

$r^2 = 9 + 16 - 36 - 40 + 61$

$r^2 = 25 - 76 + 61$

$r^2 = -51 + 61$

$r^2 = 10$

Both equations give $r^2 = 10$. The radius is $r = \sqrt{10}$.

Now we have the centre $(h, k) = (3, 4)$ and $r^2 = 10$. Substitute these values into the standard equation of the circle $(x - h)^2 + (y - k)^2 = r^2$:

This is the equation of the circle in standard form.

---

Final Equation:

The equation of the circle is $(x - 3)^2 + (y - 4)^2 = 10$.

---

Alternate Form (General Form):

Expand equation (6):

$(x^2 - 6x + 9) + (y^2 - 8y + 16) = 10$

$x^2 + y^2 - 6x - 8y + 9 + 16 = 10$

$x^2 + y^2 - 6x - 8y + 25 = 10$

Move the constant term to the left side:

$x^2 + y^2 - 6x - 8y + 25 - 10 = 0$

$x^2 + y^2 - 6x - 8y + 15 = 0$

This is the equation of the circle in general form.

Given:

The circle passes through the points $P(2, 3)$ and $Q(-1, 1)$.

The centre of the circle lies on the line $x – 3y – 11 = 0$.

---

To Find:

The equation of the circle.

---

Solution:

Let the centre of the circle be $C(h, k)$ and the radius be $r$.

The standard equation of a circle is given by:

$(x - h)^2 + (y - k)^2 = r^2$

Since the centre $C(h, k)$ lies on the line $x – 3y – 11 = 0$, it must satisfy the equation of the line:

Since the circle passes through the point $P(2, 3)$, the distance from the centre $C(h, k)$ to $P(2, 3)$ is equal to the radius $r$. Using the distance formula:

$r^2 = (2 - h)^2 + (3 - k)^2$

Expanding this equation:

$r^2 = (4 - 4h + h^2) + (9 - 6k + k^2)$

Since the circle also passes through the point $Q(-1, 1)$, the distance from the centre $C(h, k)$ to $Q(-1, 1)$ is also equal to the radius $r$:

$r^2 = (-1 - h)^2 + (1 - k)^2$

Expanding this equation:

$r^2 = (1 + 2h + h^2) + (1 - 2k + k^2)$

Equating the expressions for $r^2$ from equations (2) and (3):

$h^2 + k^2 - 4h - 6k + 13 = h^2 + k^2 + 2h - 2k + 2$

Subtract $h^2 + k^2$ from both sides:

$-4h - 6k + 13 = 2h - 2k + 2$

Move all terms involving $h$ and $k$ to the left side and constant terms to the right side:

$-4h - 2h - 6k + 2k = 2 - 13$

$-6h - 4k = -11$

Multiply by -1 to make coefficients positive:

Now we have a system of two linear equations with two variables, $h$ and $k$, from equations (1) and (4):

1) $h - 3k = 11$

4) $6h + 4k = 11$

From equation (1), we can express $h$ in terms of $k$:

Substitute equation (5) into equation (4):

$6(11 + 3k) + 4k = 11$

$66 + 18k + 4k = 11$

Combine like terms:

$66 + 22k = 11$

$22k = 11 - 66$

$22k = -55$

$\displaystyle k = \frac{-55}{22}$

Simplify the fraction by dividing both numerator and denominator by 11:

$\displaystyle k = \frac{\cancel{-55}^{-5}}{\cancel{22}_{2}}$

Now substitute the value of $k$ back into equation (5) to find $h$:

$\displaystyle h = 11 + 3\left(-\frac{5}{2}\right)$

$\displaystyle h = 11 - \frac{15}{2}$

$\displaystyle h = \frac{22 - 15}{2}$

So, the centre of the circle is $(h, k) = \left(\frac{7}{2}, -\frac{5}{2}\right)$.

Now, we need to find the radius squared, $r^2$. We can use equation (2) and substitute the values of $h = \frac{7}{2}$ and $k = -\frac{5}{2}$:

$\displaystyle r^2 = \left(\frac{7}{2}\right)^2 + \left(-\frac{5}{2}\right)^2 - 4\left(\frac{7}{2}\right) - 6\left(-\frac{5}{2}\right) + 13$

$\displaystyle r^2 = \frac{49}{4} + \frac{25}{4} - \cancel{4}^{2}\left(\frac{7}{\cancel{2}_{1}}\right) - \cancel{6}^{3}\left(\frac{-5}{\cancel{2}_{1}}\right) + 13$

$\displaystyle r^2 = \frac{49}{4} + \frac{25}{4} - 14 - (-15) + 13$

$\displaystyle r^2 = \frac{49 + 25}{4} - 14 + 15 + 13$

$\displaystyle r^2 = \frac{74}{4} + 1 + 13$

$\displaystyle r^2 = \frac{37}{2} + 14$

$\displaystyle r^2 = \frac{37 + 14 \times 2}{2}$

$\displaystyle r^2 = \frac{37 + 28}{2}$

Now we have the centre $(h, k) = \left(\frac{7}{2}, -\frac{5}{2}\right)$ and $r^2 = \frac{65}{2}$. Substitute these values into the standard equation of the circle $(x - h)^2 + (y - k)^2 = r^2$:

This is the equation of the circle in standard form.

---

Final Equation:

The equation of the circle is $\left(x - \frac{7}{2}\right)^2 + \left(y + \frac{5}{2}\right)^2 = \frac{65}{2}$.

---

Alternate Form (General Form):

Expand equation (6):

$\displaystyle x^2 - 2x\left(\frac{7}{2}\right) + \left(\frac{7}{2}\right)^2 + y^2 + 2y\left(\frac{5}{2}\right) + \left(\frac{5}{2}\right)^2 = \frac{65}{2}$

$\displaystyle x^2 - 7x + \frac{49}{4} + y^2 + 5y + \frac{25}{4} = \frac{65}{2}$

Group terms and combine constants:

$\displaystyle x^2 + y^2 - 7x + 5y + \frac{49}{4} + \frac{25}{4} = \frac{65}{2}$

$\displaystyle x^2 + y^2 - 7x + 5y + \frac{74}{4} = \frac{65}{2}$

$\displaystyle x^2 + y^2 - 7x + 5y + \frac{37}{2} = \frac{65}{2}$

Multiply the entire equation by 2 to eliminate denominators:

$\displaystyle 2\left(x^2 + y^2 - 7x + 5y + \frac{37}{2}\right) = 2\left(\frac{65}{2}\right)$

$\displaystyle 2x^2 + 2y^2 - 14x + 10y + 37 = 65$

Move the constant term to the left side:

$2x^2 + 2y^2 - 14x + 10y + 37 - 65 = 0$

$2x^2 + 2y^2 - 14x + 10y - 28 = 0$

Divide the entire equation by 2 to simplify:

$x^2 + y^2 - 7x + 5y - 14 = 0$

This is the equation of the circle in general form.

Given:

The radius of the circle is $r = 5$.

The centre of the circle lies on the x-axis.

The circle passes through the point $P(2, 3)$.

---

To Find:

The equation of the circle.

---

Solution:

Let the centre of the circle be $C(h, k)$.

Since the centre lies on the x-axis, the y-coordinate of the centre is 0. Thus, the centre is $(h, 0)$.

The radius of the circle is given as $r = 5$.

The standard equation of a circle with centre $(h, k)$ and radius $r$ is given by:

$(x - h)^2 + (y - k)^2 = r^2$

Substituting the centre $(h, 0)$ and radius $r = 5$ into the equation, we get:

$(x - h)^2 + (y - 0)^2 = 5^2$

Since the circle passes through the point $P(2, 3)$, this point must satisfy the equation of the circle. Substitute $x = 2$ and $y = 3$ into equation (1):

$(2 - h)^2 + 3^2 = 25$

Expand and simplify the equation:

$4 - 4h + h^2 + 9 = 25$

$h^2 - 4h + 13 = 25$

Move the constant term to the left side to form a quadratic equation in $h$:

$h^2 - 4h + 13 - 25 = 0$

$h^2 - 4h - 12 = 0$

Now, solve this quadratic equation for $h$. We can factor the quadratic expression. We need two numbers that multiply to -12 and add up to -4. These numbers are -6 and 2.

$(h - 6)(h + 2) = 0$

This gives two possible values for $h$:

$h - 6 = 0 \implies h = 6$

$h + 2 = 0 \implies h = -2$

This means there are two possible centres for the circle, and hence two possible equations for the circle.

---

Case 1: Centre is $(h, 0) = (6, 0)$

Substitute $h = 6$ into the standard equation $(x - h)^2 + y^2 = 25$:

---

Case 2: Centre is $(h, 0) = (-2, 0)$

Substitute $h = -2$ into the standard equation $(x - h)^2 + y^2 = 25$:

$(x - (-2))^2 + y^2 = 25$

---

Final Equations:

There are two possible equations for the circle:

1. $(x - 6)^2 + y^2 = 25$

2. $(x + 2)^2 + y^2 = 25$

---

Alternate Forms (General Form):

Expand the equations:

From Case 1:

$(x - 6)^2 + y^2 = 25$

$x^2 - 12x + 36 + y^2 = 25$

$x^2 + y^2 - 12x + 36 - 25 = 0$

$x^2 + y^2 - 12x + 11 = 0$

From Case 2:

$(x + 2)^2 + y^2 = 25$

$x^2 + 4x + 4 + y^2 = 25$

$x^2 + y^2 + 4x + 4 - 25 = 0$

$x^2 + y^2 + 4x - 21 = 0$

Given:

A circle passes through the origin (0, 0).

The circle makes an intercept of length 'a' on the x-axis, which means it passes through the point (a, 0).

The circle makes an intercept of length 'b' on the y-axis, which means it passes through the point (0, b).

---

To Find:

The equation of the circle.

---

Solution:

Let the general equation of the circle be:

We are given that the circle passes through three points: (0, 0), (a, 0), and (0, b).

Step 1: Using the point (0, 0)

Since the circle passes through the origin (0, 0), these coordinates must satisfy the equation (i).

Substituting $x = 0$ and $y = 0$ in equation (i):

$0^2 + 0^2 + 2g(0) + 2f(0) + c = 0$

$0 + 0 + 0 + 0 + c = 0$

$c = 0$

Step 2: Using the point (a, 0)

Since the circle passes through (a, 0), we substitute $x = a$, $y = 0$, and $c = 0$ in equation (i).

$a^2 + 0^2 + 2g(a) + 2f(0) + 0 = 0$

$a^2 + 2ga = 0$

$a(a + 2g) = 0$

Since $a$ is an intercept, $a \neq 0$. Therefore, $a + 2g = 0$, which gives:

$2g = -a \implies g = -\frac{a}{2}$

Step 3: Using the point (0, b)

Since the circle passes through (0, b), we substitute $x = 0$, $y = b$, and $c = 0$ in equation (i).

$0^2 + b^2 + 2g(0) + 2f(b) + 0 = 0$

$b^2 + 2fb = 0$

$b(b + 2f) = 0$

Since $b$ is an intercept, $b \neq 0$. Therefore, $b + 2f = 0$, which gives:

$2f = -b \implies f = -\frac{b}{2}$

Step 4: Finding the equation

Now, substitute the values of $g = -\frac{a}{2}$, $f = -\frac{b}{2}$, and $c = 0$ back into the general equation (i).

$x^2 + y^2 + 2\left(-\frac{a}{2}\right)x + 2\left(-\frac{b}{2}\right)y + 0 = 0$

$x^2 + y^2 - ax - by = 0$

---

Therefore, the required equation of the circle is $x^2 + y^2 - ax - by = 0$.

---

Alternate Solution

The circle passes through the points O(0, 0), A(a, 0), and B(0, b).

These three points form the vertices of a right-angled triangle OAB, with the right angle at the origin O($\angle AOB = 90^\circ$).

A property of circles states that the angle subtended by a diameter at any point on the circumference is a right angle ($90^\circ$).

Since $\angle AOB = 90^\circ$, the line segment AB connecting the points A(a, 0) and B(0, b) must be the diameter of the circle.

We know that the equation of a circle with the endpoints of a diameter at $(x_1, y_1)$ and $(x_2, y_2)$ is given by:

$(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$

Here, the endpoints of the diameter are A(a, 0) and B(0, b).

So, $x_1 = a$, $y_1 = 0$ and $x_2 = 0$, $y_2 = b$.

Substituting these values into the formula:

$(x - a)(x - 0) + (y - 0)(y - b) = 0$

$x(x - a) + y(y - b) = 0$

Expanding the terms:

$x^2 - ax + y^2 - by = 0$

Rearranging the terms, we get:

$x^2 + y^2 - ax - by = 0$

This is the same equation obtained by the first method.

Given:

The centre of the circle is at $(h, k) = (2, 2)$.

The circle passes through the point $P(4, 5)$.

---

To Find:

The equation of the circle.

---

Solution:

Let the centre of the circle be $C(2, 2)$.

The standard equation of a circle with centre $(h, k)$ and radius $r$ is given by:

$(x - h)^2 + (y - k)^2 = r^2$

Substitute the centre $(h, k) = (2, 2)$ into the equation:

Since the circle passes through the point $P(4, 5)$, the distance from the centre $C(2, 2)$ to $P(4, 5)$ is equal to the radius $r$. We can find $r^2$ by substituting the coordinates of point P $(x=4, y=5)$ into equation (1):

$(4 - 2)^2 + (5 - 2)^2 = r^2$

Simplify the equation:

$(2)^2 + (3)^2 = r^2$

$4 + 9 = r^2$

Now substitute the value of $r^2 = 13$ back into equation (1):

This is the required equation of the circle in standard form.

---

Alternate Form (General Form):

Expand equation (2):

$(x^2 - 4x + 4) + (y^2 - 4y + 4) = 13$

$x^2 + y^2 - 4x - 4y + 4 + 4 = 13$

$x^2 + y^2 - 4x - 4y + 8 = 13$

Move the constant term to the left side:

$x^2 + y^2 - 4x - 4y + 8 - 13 = 0$

$x^2 + y^2 - 4x - 4y - 5 = 0$

This is the equation of the circle in general form.

---

The equation of the circle can be written in standard form as $(x - 2)^2 + (y - 2)^2 = 13$ or in general form as $x^2 + y^2 - 4x - 4y - 5 = 0$.

Given:

The equation of the circle is $x^2 + y^2 = 25$.

The point is $P(-2.5, 3.5)$.

---

To Determine:

Whether the point $(-2.5, 3.5)$ lies inside, outside, or on the circle $x^2 + y^2 = 25$.

---

Solution:

The equation of the circle is $x^2 + y^2 = 25$. This is in the standard form with the centre at the origin $(0, 0)$ and radius $r$ such that $r^2 = 25$, so the radius is $r = \sqrt{25} = 5$.

To determine the position of a point $(x_1, y_1)$ with respect to a circle $x^2 + y^2 = r^2$, we calculate the value of $x_1^2 + y_1^2$ and compare it with $r^2$.

• If $x_1^2 + y_1^2 < r^2$, the point lies inside the circle.
• If $x_1^2 + y_1^2 = r^2$, the point lies on the circle.
• If $x_1^2 + y_1^2 > r^2$, the point lies outside the circle.

In this case, the point is $(x_1, y_1) = (-2.5, 3.5)$ and $r^2 = 25$.

Let's calculate $x_1^2 + y_1^2$ for the given point:

$\displaystyle x_1^2 + y_1^2 = (-2.5)^2 + (3.5)^2$

Calculate the squares:

$(-2.5)^2 = (-2.5) \times (-2.5) = 6.25$

$(3.5)^2 = (3.5) \times (3.5) = 12.25$

So,

$x_1^2 + y_1^2 = 6.25 + 12.25$

Now, compare this value with $r^2 = 25$:

$18.50$ compared to $25$.

Since $18.50 < 25$, the point $(-2.5, 3.5)$ lies inside the circle $x^2 + y^2 = 25$.

---

Conclusion:

The point $(-2.5, 3.5)$ lies inside the circle $x^2 + y^2 = 25$.

Given:

The equation of the parabola is $y^2 = 8x$.

---

To Find:

The coordinates of the focus, the axis, the equation of the directrix, and the length of the latus rectum.

---

Solution:

The given equation of the parabola is:

This equation is of the standard form of a parabola that opens to the right:

On comparing equation (i) and (ii), we get:

$4a = 8$

$\implies a = \frac{8}{4}$

$\implies a = 2$

Now, we can determine the properties of the parabola:

1. Coordinates of the Focus:

For a parabola of the form $y^2 = 4ax$, the focus is at $(a, 0)$.

Substituting $a=2$, the focus is (2, 0).

2. Axis of the Parabola:

The parabola $y^2 = 8x$ is symmetric about the x-axis. Therefore, the axis of the parabola is the x-axis, whose equation is y = 0.

3. Equation of the Directrix:

For a parabola of the form $y^2 = 4ax$, the equation of the directrix is $x = -a$.

Substituting $a=2$, the equation of the directrix is $x = -2$, or x + 2 = 0.

4. Length of the Latus Rectum:

The latus rectum is a line segment perpendicular to the axis of symmetry, through the focus. Its length is $4a$.

Substituting $a=2$, the length of the latus rectum is $4(2) = 8$.

---

Final Answer:

For the parabola $y^2=8x$:

• Focus: (2, 0)
• Axis: y = 0
• Equation of the Directrix: x = -2
• Length of the Latus Rectum: 8

Given:

The focus of the parabola is $F(2, 0)$.

The equation of the directrix is $x = -2$.

---

To Find:

The equation of the parabola.

---

Solution:

The vertex of a parabola is the midpoint between its focus and the point on the directrix that lies on the axis of symmetry.

The focus is at $(2, 0)$. The directrix is $x = -2$. The axis of symmetry is the line passing through the focus and perpendicular to the directrix, which is the x-axis ($y=0$).

The vertex is the midpoint of the focus $(2,0)$ and the point $(-2,0)$ on the directrix.

Vertex = $(\frac{2 + (-2)}{2}, \frac{0+0}{2}) = (0, 0)$.

Since the focus $(2, 0)$ is on the positive x-axis and the vertex is at the origin, the parabola opens to the right. The standard equation for such a parabola is:

Here, '$a$' is the distance from the vertex to the focus, which is the distance between $(0,0)$ and $(2,0)$.

$a = \sqrt{(2-0)^2 + (0-0)^2} = 2$.

Substituting $a=2$ into equation (i):

$y^2 = 4(2)x$

$\implies y^2 = 8x$

This is the required equation of the parabola.

---

Alternate Solution (Using the definition of a parabola):

A parabola is the locus of all points $P(x, y)$ that are equidistant from the focus and the directrix.

Let $P(x, y)$ be any point on the parabola. The distance from $P$ to the focus $F(2, 0)$ is $PF$.

$PF = \sqrt{(x - 2)^2 + (y - 0)^2}$

The distance from $P(x, y)$ to the directrix $x = -2$ (or $x+2=0$) is $PD$.

$PD = \frac{|x+2|}{\sqrt{1^2 + 0^2}} = |x + 2|$

By the definition of a parabola, $PF = PD$.

$\sqrt{(x - 2)^2 + y^2} = |x + 2|$

Squaring both sides:

$(x - 2)^2 + y^2 = (x + 2)^2$

Expanding the terms:

$x^2 - 4x + 4 + y^2 = x^2 + 4x + 4$

Canceling $x^2$ and $4$ from both sides:

$-4x + y^2 = 4x$

$\implies y^2 = 4x + 4x$

$\implies y^2 = 8x$

Given:

The vertex of the parabola is at $V(0, 0)$.

The focus of the parabola is at $F(0, 2)$.

---

To Find:

The equation of the parabola.

---

Solution:

Since the vertex is at the origin $(0, 0)$ and the focus is at $(0, 2)$, the focus lies on the positive y-axis.

A parabola with its vertex at the origin and focus on the y-axis is symmetric about the y-axis. As the focus is on the positive side, the parabola opens upwards.

The standard equation for an upward-opening parabola with vertex at the origin is:

The focus for this type of parabola is given by $(0, a)$.

Comparing the given focus $(0, 2)$ with the standard form $(0, a)$, we get:

$a = 2$

Now, substitute the value of $a=2$ into the standard equation (i):

$x^2 = 4(2)y$

$\implies x^2 = 8y$

This is the required equation of the parabola.

Given:

The parabola is symmetric about the y-axis.

It passes through the point $(2, -3)$.

---

To Find:

The equation of the parabola.

---

Solution:

A parabola symmetric about the y-axis has its vertex at the origin $(0,0)$ (unless shifted). Its equation is either $x^2 = 4ay$ (opens upwards) or $x^2 = -4ay$ (opens downwards).

We are given that the parabola passes through the point $(2, -3)$. This point lies in the fourth quadrant, as its y-coordinate is negative.

For a parabola symmetric about the y-axis to pass through a point with a negative y-coordinate, it must open downwards.

Therefore, the standard form of the equation must be:

Since the point $(2, -3)$ lies on this parabola, its coordinates must satisfy the equation. Substitute $x=2$ and $y=-3$ into equation (i):

$(2)^2 = -4a(-3)$

$4 = 12a$

Solving for $a$:

$a = \frac{4}{12} = \frac{1}{3}$

Now, substitute this value of $a$ back into the standard equation (i):

$x^2 = -4 \left(\frac{1}{3}\right) y$

$\implies x^2 = -\frac{4}{3}y$

This is the required equation of the parabola. It can also be written as $3x^2 = -4y$ or $3x^2 + 4y = 0$.

---

Final Answer:

The equation of the parabola is $x^2 = -\frac{4}{3}y$.

Given:

The equation of the parabola is $y^2 = 12x$.

---

To Find:

The coordinates of the focus, the axis of the parabola, the equation of the directrix, and the length of the latus rectum.

---

Solution:

The given equation of the parabola is:

This equation is of the standard form of a parabola that opens to the right with its vertex at the origin:

Comparing equation (i) with the standard form (ii), we get:

$4a = 12$

$\implies a = \frac{12}{4}$

$\implies a = 3$

Now, we can find the required properties:

1. Coordinates of the focus:

For a parabola of the form $y^2 = 4ax$, the focus is at $(a, 0)$.

Substituting $a = 3$, the focus is (3, 0).

2. Axis of the parabola:

The parabola is symmetric about the x-axis. Therefore, the axis of the parabola is the x-axis, and its equation is y = 0.

3. Equation of the directrix:

The equation of the directrix for this form is $x = -a$.

Substituting $a = 3$, the equation of the directrix is $x = -3$, or x + 3 = 0.

4. Length of the latus rectum:

The length of the latus rectum is given by $4a$.

Substituting $a = 3$, the length is $4(3) = 12$.

---

Summary of Results:

• Focus: $(3, 0)$
• Axis: $y = 0$
• Equation of the Directrix: $x = -3$
• Length of the Latus Rectum: 12

Given:

The equation of the parabola is $x^2 = 6y$.

---

To Find:

The coordinates of the focus, the axis of the parabola, the equation of the directrix, and the length of the latus rectum.

---

Solution:

The given equation of the parabola is:

This equation is of the standard form of a parabola that opens upwards with its vertex at the origin:

Comparing equation (i) with the standard form (ii), we get:

$4a = 6$

$\implies a = \frac{6}{4} = \frac{3}{2}$

Now, we can find the required properties:

1. Coordinates of the focus:

For a parabola of the form $x^2 = 4ay$, the focus is at $(0, a)$.

Substituting $a = \frac{3}{2}$, the focus is $\left(0, \frac{3}{2}\right)$.

2. Axis of the parabola:

The parabola is symmetric about the y-axis. Therefore, the axis of the parabola is the y-axis, and its equation is x = 0.

3. Equation of the directrix:

The equation of the directrix for this form is $y = -a$.

Substituting $a = \frac{3}{2}$, the equation of the directrix is $y = -\frac{3}{2}$, or 2y + 3 = 0.

4. Length of the latus rectum:

The length of the latus rectum is given by $4a$.

The length is $4 \left(\frac{3}{2}\right) = 6$.

---

Summary of Results:

• Focus: $\left(0, \frac{3}{2}\right)$
• Axis: $x = 0$
• Equation of the Directrix: $y = -\frac{3}{2}$
• Length of the Latus Rectum: 6

Given:

The equation of the parabola is $y^2 = -8x$.

---

To Find:

The coordinates of the focus, the axis of the parabola, the equation of the directrix, and the length of the latus rectum.

---

Solution:

The given equation of the parabola is:

This equation is of the standard form of a parabola that opens to the left with its vertex at the origin:

Comparing equation (i) with the standard form (ii), we get:

$-4a = -8$

$\implies a = \frac{-8}{-4}$

$\implies a = 2$

Now, we can find the required properties:

1. Coordinates of the focus:

For a parabola of the form $y^2 = -4ax$, the focus is at $(-a, 0)$.

Substituting $a = 2$, the focus is (-2, 0).

2. Axis of the parabola:

The parabola is symmetric about the x-axis. Therefore, the axis of the parabola is the x-axis, and its equation is y = 0.

3. Equation of the directrix:

The equation of the directrix for this form is $x = a$.

Substituting $a = 2$, the equation of the directrix is $x = 2$, or x - 2 = 0.

4. Length of the latus rectum:

The length of the latus rectum is given by $4a$.

Substituting $a = 2$, the length is $4(2) = 8$.

---

Summary of Results:

• Focus: $(-2, 0)$
• Axis: $y = 0$
• Equation of the Directrix: $x = 2$
• Length of the Latus Rectum: 8

Given:

The equation of the parabola is $x^2 = -16y$.

---

To Find:

The coordinates of the focus, the axis of the parabola, the equation of the directrix, and the length of the latus rectum.

---

Solution:

The given equation of the parabola is:

This equation is of the standard form of a parabola that opens downwards with its vertex at the origin:

Comparing equation (i) with the standard form (ii), we get:

$-4a = -16$

$\implies a = \frac{-16}{-4}$

$\implies a = 4$

Now, we can find the required properties:

1. Coordinates of the focus:

For a parabola of the form $x^2 = -4ay$, the focus is at $(0, -a)$.

Substituting $a = 4$, the focus is (0, -4).

2. Axis of the parabola:

The parabola is symmetric about the y-axis. Therefore, the axis of the parabola is the y-axis, and its equation is x = 0.

3. Equation of the directrix:

The equation of the directrix for this form is $y = a$.

Substituting $a = 4$, the equation of the directrix is $y = 4$, or y - 4 = 0.

4. Length of the latus rectum:

The length of the latus rectum is given by $4a$.

Substituting $a = 4$, the length is $4(4) = 16$.

---

Summary of Results:

• Focus: $(0, -4)$
• Axis: $x = 0$
• Equation of the Directrix: $y = 4$
• Length of the Latus Rectum: 16

Given:

The equation of the parabola is $y^2 = 10x$.

---

To Find:

The coordinates of the focus, the axis of the parabola, the equation of the directrix, and the length of the latus rectum.

---

Solution:

The given equation of the parabola is:

This equation is of the standard form of a parabola that opens to the right with its vertex at the origin:

Comparing equation (i) with the standard form (ii), we get:

$4a = 10$

$\implies a = \frac{10}{4} = \frac{5}{2}$

Now, we can find the required properties:

1. Coordinates of the focus:

For a parabola of the form $y^2 = 4ax$, the focus is at $(a, 0)$.

Substituting $a = \frac{5}{2}$, the focus is $\left(\frac{5}{2}, 0\right)$.

2. Axis of the parabola:

The parabola is symmetric about the x-axis. Therefore, the axis of the parabola is the x-axis, and its equation is y = 0.

3. Equation of the directrix:

The equation of the directrix for this form is $x = -a$.

Substituting $a = \frac{5}{2}$, the equation is $x = -\frac{5}{2}$, or 2x + 5 = 0.

4. Length of the latus rectum:

The length of the latus rectum is given by $4a$.

The length is $4\left(\frac{5}{2}\right) = 10$.

---

Summary of Results:

• Focus: $\left(\frac{5}{2}, 0\right)$
• Axis: $y = 0$
• Equation of the Directrix: $x = -\frac{5}{2}$
• Length of the Latus Rectum: 10

Given:

The equation of the parabola is $x^2 = -9y$.

---

To Find:

The coordinates of the focus, the axis of the parabola, the equation of the directrix, and the length of the latus rectum.

---

Solution:

The given equation of the parabola is:

This equation is of the standard form of a parabola that opens downwards with its vertex at the origin:

Comparing equation (i) with the standard form (ii), we get:

$-4a = -9$

$\implies a = \frac{-9}{-4} = \frac{9}{4}$

Now, we can find the required properties:

1. Coordinates of the focus:

For a parabola of the form $x^2 = -4ay$, the focus is at $(0, -a)$.

Substituting $a = \frac{9}{4}$, the focus is $\left(0, -\frac{9}{4}\right)$.

2. Axis of the parabola:

The parabola is symmetric about the y-axis. Therefore, the axis of the parabola is the y-axis, and its equation is x = 0.

3. Equation of the directrix:

The equation of the directrix for this form is $y = a$.

Substituting $a = \frac{9}{4}$, the equation is $y = \frac{9}{4}$, or 4y - 9 = 0.

4. Length of the latus rectum:

The length of the latus rectum is given by $4a$.

The length is $4\left(\frac{9}{4}\right) = 9$.

---

Summary of Results:

• Focus: $\left(0, -\frac{9}{4}\right)$
• Axis: $x = 0$
• Equation of the Directrix: $y = \frac{9}{4}$
• Length of the Latus Rectum: 9

Given:

The focus of the parabola is $F(6, 0)$.

The directrix of the parabola is the line $x = -6$.

---

To Find:

The equation of the parabola.

---

Solution:

The focus is at $(6, 0)$ and the directrix is the line $x = -6$.

Since the focus lies on the x-axis and the directrix is a vertical line, the axis of symmetry of the parabola is the x-axis ($y=0$).

The vertex of the parabola is the midpoint between the focus and the directrix. The point on the directrix corresponding to the focus is $(-6, 0)$.

Vertex = $\left( \frac{6 + (-6)}{2}, \frac{0 + 0}{2} \right) = (0, 0)$.

Since the focus $(6, 0)$ is to the right of the vertex $(0, 0)$, the parabola opens to the right. The standard equation for such a parabola is:

Here, '$a$' is the distance from the vertex to the focus. The distance between $(0,0)$ and $(6,0)$ is $a=6$.

Substituting the value of $a = 6$ into equation (i):

$y^2 = 4(6)x$

$\implies y^2 = 24x$

This is the required equation of the parabola.

---

Alternate Method (Using the definition of a parabola):

By definition, a parabola is the locus of a point $P(x, y)$ that is equidistant from the focus $F(6, 0)$ and the directrix $x + 6 = 0$.

Distance from P to F: $PF = \sqrt{(x - 6)^2 + (y - 0)^2}$

Distance from P to the directrix: $PD = |x + 6|$

Setting $PF = PD$:

$\sqrt{(x - 6)^2 + y^2} = |x + 6|$

Squaring both sides:

$(x - 6)^2 + y^2 = (x + 6)^2$

$x^2 - 12x + 36 + y^2 = x^2 + 12x + 36$

Canceling common terms ($x^2$ and 36) from both sides:

$-12x + y^2 = 12x$

$\implies y^2 = 24x$

Both methods yield the same result.

---

Final Answer:

The equation of the parabola is $y^2 = 24x$.

Given:

The focus of the parabola is $F(0, -3)$.

The directrix of the parabola is the line $y = 3$.

---

To Find:

The equation of the parabola.

---

Solution:

The focus is at $(0, -3)$ and the directrix is the line $y = 3$.

Since the focus lies on the y-axis and the directrix is a horizontal line, the axis of symmetry of the parabola is the y-axis ($x=0$).

The vertex is the midpoint of the focus $(0,-3)$ and the corresponding point on the directrix $(0,3)$.

Vertex = $\left( \frac{0 + 0}{2}, \frac{-3 + 3}{2} \right) = (0, 0)$.

Since the focus $(0, -3)$ is below the vertex $(0, 0)$, the parabola opens downwards. The standard equation for such a parabola is:

Here, '$a$' is the distance from the vertex to the focus. The distance between $(0,0)$ and $(0,-3)$ is $a=3$.

Substituting the value of $a = 3$ into equation (i):

$x^2 = -4(3)y$

$\implies x^2 = -12y$

This is the required equation of the parabola.

---

Alternate Method (Using the definition of a parabola):

Let $P(x, y)$ be any point on the parabola. Its distance from the focus $F(0, -3)$ must equal its distance from the directrix $y - 3 = 0$.

Distance from P to F: $PF = \sqrt{(x - 0)^2 + (y - (-3))^2} = \sqrt{x^2 + (y + 3)^2}$

Distance from P to the directrix: $PD = |y - 3|$

Setting $PF = PD$:

$\sqrt{x^2 + (y + 3)^2} = |y - 3|$

Squaring both sides:

$x^2 + (y + 3)^2 = (y - 3)^2$

$x^2 + y^2 + 6y + 9 = y^2 - 6y + 9$

Canceling common terms ($y^2$ and 9) from both sides:

$x^2 + 6y = -6y$

$\implies x^2 = -12y$

---

Final Answer:

The equation of the parabola is $x^2 = -12y$.

Given:

The vertex of the parabola is at $(0, 0)$.

The focus of the parabola is at $(3, 0)$.

---

To Find:

The equation of the parabola.

---

Solution:

The vertex is at the origin $(0, 0)$ and the focus is at $(3, 0)$.

Since the focus lies on the positive x-axis, the parabola is symmetric about the x-axis and opens to the right.

The standard equation for such a parabola is:

The focus is at $(a, 0)$. Comparing this with the given focus $(3, 0)$, we get:

$a = 3$

Substituting $a = 3$ into equation (i):

$y^2 = 4(3)x$

$\implies y^2 = 12x$

This is the required equation of the parabola.

---

Final Answer:

The equation of the parabola is $y^2 = 12x$.

Given:

The vertex of the parabola is at $(0, 0)$.

The focus of the parabola is at $(-2, 0)$.

---

To Find:

The equation of the parabola.

---

Solution:

The vertex is at the origin $(0, 0)$ and the focus is at $(-2, 0)$.

Since the focus lies on the negative x-axis, the parabola is symmetric about the x-axis and opens to the left.

The standard equation for such a parabola is:

The focus is at $(-a, 0)$. Comparing this with the given focus $(-2, 0)$, we get:

$a = 2$

Substituting $a = 2$ into equation (i):

$y^2 = -4(2)x$

$\implies y^2 = -8x$

This is the required equation of the parabola.

---

Final Answer:

The equation of the parabola is $y^2 = -8x$.

Given:

The vertex of the parabola is at $(0, 0)$.

The parabola passes through the point $(2, 3)$.

The axis of symmetry is the x-axis.

---

To Find:

The equation of the parabola.

---

Solution:

Since the vertex is at $(0,0)$ and the axis of symmetry is the x-axis, the equation of the parabola is of the form $y^2 = 4ax$ or $y^2 = -4ax$.

The parabola passes through the point $(2, 3)$, which lies in the first quadrant. Since the x-coordinate is positive, the parabola must open to the right.

Therefore, the equation of the parabola is of the form:

Since the point $(2, 3)$ lies on the parabola, it must satisfy the equation. We substitute $x = 2$ and $y = 3$ into equation (i):

$(3)^2 = 4a(2)$

$9 = 8a$

$\implies a = \frac{9}{8}$

Now, substitute the value of $a$ back into the standard equation (i):

$y^2 = 4\left(\frac{9}{8}\right)x$

$y^2 = \frac{36}{8}x$

$\implies y^2 = \frac{9}{2}x$

This is the required equation of the parabola.

---

Final Answer:

The equation of the parabola is $y^2 = \frac{9}{2}x$ or $2y^2 = 9x$.

Given:

The vertex of the parabola is at $(0, 0)$.

The parabola passes through the point $(5, 2)$.

The parabola is symmetric with respect to the y-axis.

---

To Find:

The equation of the parabola.

---

Solution:

Since the vertex is at $(0,0)$ and the parabola is symmetric with respect to the y-axis, its equation is of the form $x^2 = 4ay$ or $x^2 = -4ay$.

The parabola passes through the point $(5, 2)$, which lies in the first quadrant. Since the y-coordinate is positive, the parabola must open upwards.

Therefore, the equation of the parabola is of the form:

Since the point $(5, 2)$ lies on the parabola, its coordinates must satisfy the equation. We substitute $x = 5$ and $y = 2$ into equation (i):

$(5)^2 = 4a(2)$

$25 = 8a$

$\implies a = \frac{25}{8}$

Now, substitute the value of $a$ back into the standard equation (i):

$x^2 = 4\left(\frac{25}{8}\right)y$

$x^2 = \frac{100}{8}y$

$\implies x^2 = \frac{25}{2}y$

This is the required equation of the parabola.

---

Final Answer:

The equation of the parabola is $x^2 = \frac{25}{2}y$ or $2x^2 = 25y$.

Given:

The equation of the ellipse is $\frac{x^{2}}{25} + \frac{y^{2}}{9} = 1$.

---

To Find:

The coordinates of the foci and vertices, the length of the major axis and minor axis, the eccentricity, and the length of the latus rectum.

---

Solution:

The given equation of the ellipse is:

The standard equation of an ellipse with its centre at the origin is $\frac{x^{2}}{a^2} + \frac{y^{2}}{b^2} = 1$.

Comparing equation (i) with the standard form, we get:

$a^2 = 25 \implies a = 5$

$b^2 = 9 \implies b = 3$

Since the denominator of the $x^2$ term is larger than the denominator of the $y^2$ term ($a^2 > b^2$), the major axis is along the x-axis.

The relationship between $a$, $b$, and $c$ (the distance from the centre to a focus) is given by $c^2 = a^2 - b^2$.

$c^2 = 25 - 9 = 16$

$\implies c = \sqrt{16} = 4$

Now we can determine the properties of the ellipse:

1. Coordinates of the foci: The foci are located at $(\pm c, 0)$. Thus, the foci are $(\pm 4, 0)$.

2. Coordinates of the vertices: The vertices are located at $(\pm a, 0)$. Thus, the vertices are $(\pm 5, 0)$.

3. Length of the major axis: The length is $2a = 2(5) = 10$.

4. Length of the minor axis: The length is $2b = 2(3) = 6$.

5. Eccentricity ($e$): The eccentricity is given by the formula $e = \frac{c}{a}$.

$e = \frac{4}{5}$

6. Length of the latus rectum: The length is given by the formula $\frac{2b^2}{a}$.

Length = $\frac{2(9)}{5} = \frac{18}{5}$.

---

Summary of Results:

• Foci: $(\pm 4, 0)$
• Vertices: $(\pm 5, 0)$
• Length of Major Axis: 10
• Length of Minor Axis: 6
• Eccentricity: $\frac{4}{5}$
• Length of Latus Rectum: $\frac{18}{5}$

Given:

The equation of the ellipse is $9x^2 + 4y^2 = 36$.

---

To Find:

The coordinates of the foci and vertices, the lengths of the major and minor axes, and the eccentricity.

---

Solution:

The given equation is $9x^2 + 4y^2 = 36$.

To convert this into the standard form of an ellipse, we divide the entire equation by 36 to make the right-hand side equal to 1.

$\frac{9x^2}{36} + \frac{4y^2}{36} = \frac{36}{36}$

Since the denominator of the $y^2$ term (9) is larger than the denominator of the $x^2$ term (4), the major axis is along the y-axis. The standard form for this type of ellipse is $\frac{x^{2}}{b^2} + \frac{y^{2}}{a^2} = 1$.

Comparing equation (i) with the standard form, we get:

$a^2 = 9 \implies a = 3$

$b^2 = 4 \implies b = 2$

The relationship between $a$, $b$, and $c$ is $c^2 = a^2 - b^2$.

$c^2 = 9 - 4 = 5$

$\implies c = \sqrt{5}$

Now we can determine the properties of the ellipse:

1. Coordinates of the foci: The foci are located at $(0, \pm c)$. Thus, the foci are $(0, \pm \sqrt{5})$.

2. Coordinates of the vertices: The vertices are located at $(0, \pm a)$. Thus, the vertices are $(0, \pm 3)$.

3. Length of the major axis: The length is $2a = 2(3) = 6$.

4. Length of the minor axis: The length is $2b = 2(2) = 4$.

5. Eccentricity ($e$): The eccentricity is $e = \frac{c}{a} = \frac{\sqrt{5}}{3}$.

---

Summary of Results:

• Foci: $(0, \pm \sqrt{5})$
• Vertices: $(0, \pm 3)$
• Length of Major Axis: 6
• Length of Minor Axis: 4
• Eccentricity: $\frac{\sqrt{5}}{3}$

Given:

The vertices of the ellipse are $(\pm 13, 0)$.

The foci of the ellipse are $(\pm 5, 0)$.

---

To Find:

The equation of the ellipse.

---

Solution:

The vertices $(\pm 13, 0)$ and foci $(\pm 5, 0)$ both lie on the x-axis. This implies that the major axis is along the x-axis and the centre of the ellipse is at the origin $(0, 0)$.

The standard equation for this type of ellipse is:

The coordinates of the vertices are $(\pm a, 0)$. From the given vertices $(\pm 13, 0)$, we have:

$a = 13 \implies a^2 = 169$

The coordinates of the foci are $(\pm c, 0)$. From the given foci $(\pm 5, 0)$, we have:

$c = 5 \implies c^2 = 25$

The relationship between $a$, $b$, and $c$ for an ellipse is $c^2 = a^2 - b^2$. We can find $b^2$ as follows:

$b^2 = a^2 - c^2$

$b^2 = 169 - 25$

$b^2 = 144$

Now, substitute the values of $a^2 = 169$ and $b^2 = 144$ into the standard equation (i):

$\frac{x^2}{169} + \frac{y^2}{144} = 1$

---

Final Answer:

The required equation of the ellipse is $\frac{x^2}{169} + \frac{y^2}{144} = 1$.

Given:

The length of the major axis is 20.

The foci of the ellipse are $(0, \pm 5)$.

---

To Find:

The equation of the ellipse.

---

Solution:

The foci are at $(0, \pm 5)$, which means they lie on the y-axis. Therefore, the major axis is along the y-axis and the centre of the ellipse is at the origin $(0, 0)$.

The standard equation for this type of ellipse is:

The length of the major axis is $2a$. We are given:

$2a = 20 \implies a = 10$

So, $a^2 = 100$.

The coordinates of the foci are $(0, \pm c)$. From the given foci $(0, \pm 5)$, we have:

$c = 5 \implies c^2 = 25$

The relationship between $a$, $b$, and $c$ is $c^2 = a^2 - b^2$. We can find $b^2$ as follows:

$b^2 = a^2 - c^2$

$b^2 = 100 - 25$

$b^2 = 75$

Now, substitute the values of $a^2 = 100$ and $b^2 = 75$ into the standard equation (i):

$\frac{x^2}{75} + \frac{y^2}{100} = 1$

---

Final Answer:

The required equation of the ellipse is $\frac{x^2}{75} + \frac{y^2}{100} = 1$.

Given:

The major axis of the ellipse is along the x-axis.

The ellipse passes through the points $(4, 3)$ and $(-1, 4)$.

---

To Find:

The equation of the ellipse.

---

Solution:

Since the major axis is along the x-axis and no other information about the centre is given, we assume the centre is at the origin $(0,0)$.

The standard equation of the ellipse is:

Since the ellipse passes through the point $(4, 3)$, these coordinates must satisfy the equation:

$\frac{(4)^2}{a^2} + \frac{(3)^2}{b^2} = 1$

The ellipse also passes through the point $(-1, 4)$, so these coordinates must also satisfy the equation:

$\frac{(-1)^2}{a^2} + \frac{(4)^2}{b^2} = 1$

We now have a system of two equations with two variables, $a^2$ and $b^2$.

From equation (iii), we can express $\frac{1}{a^2}$ in terms of $\frac{1}{b^2}$:

Substitute this expression into equation (ii):

$16 \left( 1 - \frac{16}{b^2} \right) + \frac{9}{b^2} = 1$

$16 - \frac{256}{b^2} + \frac{9}{b^2} = 1$

$16 - \frac{247}{b^2} = 1$

$15 = \frac{247}{b^2}$

$\implies b^2 = \frac{247}{15}$

Now substitute the value of $b^2$ back into equation (iv) to find $a^2$.

$\frac{1}{a^2} = 1 - \frac{16}{(247/15)}$

$\frac{1}{a^2} = 1 - \frac{16 \times 15}{247} = 1 - \frac{240}{247}$

$\frac{1}{a^2} = \frac{247 - 240}{247} = \frac{7}{247}$

$\implies a^2 = \frac{247}{7}$

We have $a^2 = \frac{247}{7}$ and $b^2 = \frac{247}{15}$. Substituting these values into the standard equation (i):

$\frac{x^2}{(247/7)} + \frac{y^2}{(247/15)} = 1$

$\implies \frac{7x^2}{247} + \frac{15y^2}{247} = 1$

Multiplying both sides by 247 gives the final equation:

$7x^2 + 15y^2 = 247$

---

Final Answer:

The required equation of the ellipse is $7x^2 + 15y^2 = 247$.

Given:

The equation of the ellipse is $\frac{x^{2}}{36} + \frac{y^{2}}{16} = 1$.

---

To Find:

The coordinates of the foci, the vertices, the length of the major axis, the length of the minor axis, the eccentricity, and the length of the latus rectum of the given ellipse.

---

Solution:

The given equation of the ellipse is:

This equation is in the standard form $\frac{x^{2}}{a^2} + \frac{y^{2}}{b^2} = 1$, where the centre of the ellipse is at the origin $(0, 0)$.

Comparing equation (i) with the standard form, we have:

$a^2 = 36 \implies a = \sqrt{36} = 6$ (since $a > 0$)

$b^2 = 16 \implies b = \sqrt{16} = 4$ (since $b > 0$)

Since $a^2 = 36 > b^2 = 16$, the major axis is along the x-axis.

Now we can find the required properties of the ellipse:

1. Coordinates of the foci: For an ellipse with the major axis along the x-axis, the coordinates of the foci are $(\pm c, 0)$, where $c^2 = a^2 - b^2$.

Calculate $c^2$:

$c^2 = 36 - 16$

$c^2 = 20$

$c = \sqrt{20} = \sqrt{4 \times 5} = 2\sqrt{5}$ (since $c > 0$)

The coordinates of the foci are $(\pm 2\sqrt{5}, 0)$, i.e., $(2\sqrt{5}, 0)$ and $(-2\sqrt{5}, 0)$.

2. Vertices: For an ellipse with the major axis along the x-axis, the coordinates of the vertices are $(\pm a, 0)$.

Substitute $a = 6$, the coordinates of the vertices are $(\pm 6, 0)$, i.e., $(6, 0)$ and $(-6, 0)$.

3. Length of the major axis: The length of the major axis is $2a$.

Substitute $a = 6$, the length of the major axis is $2 \times 6 = 12$.

4. Length of the minor axis: The length of the minor axis is $2b$.

Substitute $b = 4$, the length of the minor axis is $2 \times 4 = 8$.

5. Eccentricity: The eccentricity $e$ is given by $e = \frac{c}{a}$.

Substitute $c = 2\sqrt{5}$ and $a = 6$, the eccentricity is $\displaystyle e = \frac{2\sqrt{5}}{6}$.

Simplify the fraction:

$\displaystyle e = \frac{\cancel{2}\sqrt{5}}{\cancel{6}_{3}} = \frac{\sqrt{5}}{3}$

6. Length of the latus rectum: The length of the latus rectum is $\displaystyle \frac{2b^2}{a}$.

Substitute $b^2 = 16$ and $a = 6$, the length of the latus rectum is $\displaystyle \frac{2 \times 16}{6}$.

Simplify the fraction:

$\displaystyle \frac{32}{6} = \frac{\cancel{32}^{16}}{\cancel{6}_{3}} = \frac{16}{3}$

---

Summary of Results:

Foci: $(\pm 2\sqrt{5}, 0)$

Vertices: $(\pm 6, 0)$

Length of Major Axis: 12

Length of Minor Axis: 8

Eccentricity: $\frac{\sqrt{5}}{3}$

Length of Latus Rectum: $\frac{16}{3}$

Given:

The equation of the ellipse is $\frac{x^{2}}{4} + \frac{y^{2}}{25} = 1$.

---

To Find:

The coordinates of the foci, the vertices, the length of the major axis, the length of the minor axis, the eccentricity, and the length of the latus rectum of the given ellipse.

---

Solution:

The given equation of the ellipse is:

This equation is in the standard form $\frac{x^{2}}{b^2} + \frac{y^{2}}{a^2} = 1$, where the centre of the ellipse is at the origin $(0, 0)$, because the denominator of the $y^2$ term (25) is greater than the denominator of the $x^2$ term (4).

Comparing equation (i) with the standard form $\frac{x^{2}}{b^2} + \frac{y^{2}}{a^2} = 1$, we have:

$a^2 = 25 \implies a = \sqrt{25} = 5$ (since $a > 0$)

$b^2 = 4 \implies b = \sqrt{4} = 2$ (since $b > 0$)

Since $a^2 = 25 > b^2 = 4$, the major axis is along the y-axis.

Now we can find the required properties of the ellipse with the major axis along the y-axis:

1. Coordinates of the foci: For an ellipse with the major axis along the y-axis, the coordinates of the foci are $(0, \pm c)$, where $c^2 = a^2 - b^2$.

Calculate $c^2$:

$c^2 = 25 - 4$

$c^2 = 21$

$c = \sqrt{21}$ (since $c > 0$)

The coordinates of the foci are $(0, \pm \sqrt{21})$, i.e., $(0, \sqrt{21})$ and $(0, -\sqrt{21})$.

2. Vertices: For an ellipse with the major axis along the y-axis, the coordinates of the vertices are $(0, \pm a)$.

Substitute $a = 5$, the coordinates of the vertices are $(0, \pm 5)$, i.e., $(0, 5)$ and $(0, -5)$.

3. Length of the major axis: The length of the major axis is $2a$.

Substitute $a = 5$, the length of the major axis is $2 \times 5 = 10$.

4. Length of the minor axis: The length of the minor axis is $2b$.

Substitute $b = 2$, the length of the minor axis is $2 \times 2 = 4$.

5. Eccentricity: The eccentricity $e$ is given by $e = \frac{c}{a}$.

Substitute $c = \sqrt{21}$ and $a = 5$, the eccentricity is $\displaystyle e = \frac{\sqrt{21}}{5}$.

6. Length of the latus rectum: The length of the latus rectum is $\displaystyle \frac{2b^2}{a}$.

Substitute $b^2 = 4$ and $a = 5$, the length of the latus rectum is $\displaystyle \frac{2 \times 4}{5} = \frac{8}{5}$.

---

Summary of Results:

Foci: $(0, \pm \sqrt{21})$

Vertices: $(0, \pm 5)$

Length of Major Axis: 10

Length of Minor Axis: 4

Eccentricity: $\frac{\sqrt{21}}{5}$

Length of Latus Rectum: $\frac{8}{5}$

Given:

The equation of the ellipse is $\frac{x^{2}}{16} + \frac{y^{2}}{9} = 1$.

---

To Find:

The coordinates of the foci, the vertices, the length of the major axis, the length of the minor axis, the eccentricity, and the length of the latus rectum of the given ellipse.

---

Solution:

The given equation of the ellipse is:

This equation is in the standard form $\frac{x^{2}}{a^2} + \frac{y^{2}}{b^2} = 1$, where the centre of the ellipse is at the origin $(0, 0)$, because the denominator of the $x^2$ term (16) is greater than the denominator of the $y^2$ term (9).

Comparing equation (i) with the standard form $\frac{x^{2}}{a^2} + \frac{y^{2}}{b^2} = 1$, we have:

$a^2 = 16 \implies a = \sqrt{16} = 4$ (since $a > 0$)

$b^2 = 9 \implies b = \sqrt{9} = 3$ (since $b > 0$)

Since $a^2 = 16 > b^2 = 9$, the major axis is along the x-axis.

Now we can find the required properties of the ellipse with the major axis along the x-axis:

1. Coordinates of the foci: For an ellipse with the major axis along the x-axis, the coordinates of the foci are $(\pm c, 0)$, where $c^2 = a^2 - b^2$.

Calculate $c^2$:

$c^2 = 16 - 9$

$c^2 = 7$

$c = \sqrt{7}$ (since $c > 0$)

The coordinates of the foci are $(\pm \sqrt{7}, 0)$, i.e., $(\sqrt{7}, 0)$ and $(-\sqrt{7}, 0)$.

2. Vertices: For an ellipse with the major axis along the x-axis, the coordinates of the vertices are $(\pm a, 0)$.

Substitute $a = 4$, the coordinates of the vertices are $(\pm 4, 0)$, i.e., $(4, 0)$ and $(-4, 0)$.

3. Length of the major axis: The length of the major axis is $2a$.

Substitute $a = 4$, the length of the major axis is $2 \times 4 = 8$.

4. Length of the minor axis: The length of the minor axis is $2b$.

Substitute $b = 3$, the length of the minor axis is $2 \times 3 = 6$.

5. Eccentricity: The eccentricity $e$ is given by $e = \frac{c}{a}$.

Substitute $c = \sqrt{7}$ and $a = 4$, the eccentricity is $\displaystyle e = \frac{\sqrt{7}}{4}$.

6. Length of the latus rectum: The length of the latus rectum is $\displaystyle \frac{2b^2}{a}$.

Substitute $b^2 = 9$ and $a = 4$, the length of the latus rectum is $\displaystyle \frac{2 \times 9}{4}$.

Simplify the fraction:

$\displaystyle \frac{18}{4} = \frac{\cancel{18}^{9}}{\cancel{4}_{2}} = \frac{9}{2}$

---

Summary of Results:

Foci: $(\pm \sqrt{7}, 0)$

Vertices: $(\pm 4, 0)$

Length of Major Axis: 8

Length of Minor Axis: 6

Eccentricity: $\frac{\sqrt{7}}{4}$

Length of Latus Rectum: $\frac{9}{2}$

Given:

The equation of the ellipse is $\frac{x^{2}}{25} + \frac{y^{2}}{100} = 1$.

---

To Find:

The coordinates of the foci, the vertices, the length of the major axis, the length of the minor axis, the eccentricity, and the length of the latus rectum of the given ellipse.

---

Solution:

The given equation of the ellipse is:

This equation is in the standard form $\frac{x^{2}}{b^2} + \frac{y^{2}}{a^2} = 1$, where the centre of the ellipse is at the origin $(0, 0)$, because the denominator of the $y^2$ term (100) is greater than the denominator of the $x^2$ term (25).

Comparing equation (i) with the standard form $\frac{x^{2}}{b^2} + \frac{y^{2}}{a^2} = 1$, we have:

$a^2 = 100 \implies a = \sqrt{100} = 10$ (since $a > 0$)

$b^2 = 25 \implies b = \sqrt{25} = 5$ (since $b > 0$)

Since $a^2 = 100 > b^2 = 25$, the major axis is along the y-axis.

Now we can find the required properties of the ellipse with the major axis along the y-axis:

1. Coordinates of the foci: For an ellipse with the major axis along the y-axis, the coordinates of the foci are $(0, \pm c)$, where $c^2 = a^2 - b^2$.

Calculate $c^2$:

$c^2 = 100 - 25$

$c^2 = 75$

$c = \sqrt{75} = \sqrt{25 \times 3} = 5\sqrt{3}$ (since $c > 0$)

The coordinates of the foci are $(0, \pm 5\sqrt{3})$, i.e., $(0, 5\sqrt{3})$ and $(0, -5\sqrt{3})$.

2. Vertices: For an ellipse with the major axis along the y-axis, the coordinates of the vertices are $(0, \pm a)$.

Substitute $a = 10$, the coordinates of the vertices are $(0, \pm 10)$, i.e., $(0, 10)$ and $(0, -10)$.

3. Length of the major axis: The length of the major axis is $2a$.

Substitute $a = 10$, the length of the major axis is $2 \times 10 = 20$.

4. Length of the minor axis: The length of the minor axis is $2b$.

Substitute $b = 5$, the length of the minor axis is $2 \times 5 = 10$.

5. Eccentricity: The eccentricity $e$ is given by $e = \frac{c}{a}$.

Substitute $c = 5\sqrt{3}$ and $a = 10$, the eccentricity is $\displaystyle e = \frac{5\sqrt{3}}{10}$.

Simplify the fraction:

$\displaystyle e = \frac{\cancel{5}\sqrt{3}}{\cancel{10}_{2}} = \frac{\sqrt{3}}{2}$

6. Length of the latus rectum: The length of the latus rectum is $\displaystyle \frac{2b^2}{a}$.

Substitute $b^2 = 25$ and $a = 10$, the length of the latus rectum is $\displaystyle \frac{2 \times 25}{10}$.

Simplify the fraction:

$\displaystyle \frac{50}{10} = 5$

---

Summary of Results:

Foci: $(0, \pm 5\sqrt{3})$

Vertices: $(0, \pm 10)$

Length of Major Axis: 20

Length of Minor Axis: 10

Eccentricity: $\frac{\sqrt{3}}{2}$

Length of Latus Rectum: 5

Given:

The equation of the ellipse is $\frac{x^{2}}{49} + \frac{y^{2}}{36} = 1$.

---

To Find:

The coordinates of the foci, the vertices, the length of the major axis, the length of the minor axis, the eccentricity, and the length of the latus rectum of the given ellipse.

---

Solution:

The given equation of the ellipse is:

This equation is in the standard form $\frac{x^{2}}{a^2} + \frac{y^{2}}{b^2} = 1$, where the centre of the ellipse is at the origin $(0, 0)$, because the denominator of the $x^2$ term (49) is greater than the denominator of the $y^2$ term (36).

Comparing equation (i) with the standard form $\frac{x^{2}}{a^2} + \frac{y^{2}}{b^2} = 1$, we have:

$a^2 = 49 \implies a = \sqrt{49} = 7$ (since $a > 0$)

$b^2 = 36 \implies b = \sqrt{36} = 6$ (since $b > 0$)

Since $a^2 = 49 > b^2 = 36$, the major axis is along the x-axis.

Now we can find the required properties of the ellipse with the major axis along the x-axis:

1. Coordinates of the foci: For an ellipse with the major axis along the x-axis, the coordinates of the foci are $(\pm c, 0)$, where $c^2 = a^2 - b^2$.

Calculate $c^2$:

$c^2 = 49 - 36$

$c^2 = 13$

$c = \sqrt{13}$ (since $c > 0$)

The coordinates of the foci are $(\pm \sqrt{13}, 0)$, i.e., $(\sqrt{13}, 0)$ and $(-\sqrt{13}, 0)$.

2. Vertices: For an ellipse with the major axis along the x-axis, the coordinates of the vertices are $(\pm a, 0)$.

Substitute $a = 7$, the coordinates of the vertices are $(\pm 7, 0)$, i.e., $(7, 0)$ and $(-7, 0)$.

3. Length of the major axis: The length of the major axis is $2a$.

Substitute $a = 7$, the length of the major axis is $2 \times 7 = 14$.

4. Length of the minor axis: The length of the minor axis is $2b$.

Substitute $b = 6$, the length of the minor axis is $2 \times 6 = 12$.

5. Eccentricity: The eccentricity $e$ is given by $e = \frac{c}{a}$.

Substitute $c = \sqrt{13}$ and $a = 7$, the eccentricity is $\displaystyle e = \frac{\sqrt{13}}{7}$.

6. Length of the latus rectum: The length of the latus rectum is $\displaystyle \frac{2b^2}{a}$.

Substitute $b^2 = 36$ and $a = 7$, the length of the latus rectum is $\displaystyle \frac{2 \times 36}{7} = \frac{72}{7}$.

---

Summary of Results:

Foci: $(\pm \sqrt{13}, 0)$

Vertices: $(\pm 7, 0)$

Length of Major Axis: 14

Length of Minor Axis: 12

Eccentricity: $\frac{\sqrt{13}}{7}$

Length of Latus Rectum: $\frac{72}{7}$

Given:

The equation of the ellipse is $\frac{x^{2}}{100} + \frac{y^{2}}{400} = 1$.

---

To Find:

The coordinates of the foci, the vertices, the length of the major axis, the length of the minor axis, the eccentricity, and the length of the latus rectum of the given ellipse.

---

Solution:

The given equation of the ellipse is:

This equation is in the standard form $\frac{x^{2}}{b^2} + \frac{y^{2}}{a^2} = 1$, where the centre of the ellipse is at the origin $(0, 0)$, because the denominator of the $y^2$ term (400) is greater than the denominator of the $x^2$ term (100).

Comparing equation (i) with the standard form $\frac{x^{2}}{b^2} + \frac{y^{2}}{a^2} = 1$, we have:

$a^2 = 400 \implies a = \sqrt{400} = 20$ (since $a > 0$)

$b^2 = 100 \implies b = \sqrt{100} = 10$ (since $b > 0$)

Since $a^2 = 400 > b^2 = 100$, the major axis is along the y-axis.

Now we can find the required properties of the ellipse with the major axis along the y-axis:

1. Coordinates of the foci: For an ellipse with the major axis along the y-axis, the coordinates of the foci are $(0, \pm c)$, where $c^2 = a^2 - b^2$.

Calculate $c^2$:

$c^2 = 400 - 100$

$c^2 = 300$

$c = \sqrt{300} = \sqrt{100 \times 3} = 10\sqrt{3}$ (since $c > 0$)

The coordinates of the foci are $(0, \pm 10\sqrt{3})$, i.e., $(0, 10\sqrt{3})$ and $(0, -10\sqrt{3})$.

2. Vertices: For an ellipse with the major axis along the y-axis, the coordinates of the vertices are $(0, \pm a)$.

Substitute $a = 20$, the coordinates of the vertices are $(0, \pm 20)$, i.e., $(0, 20)$ and $(0, -20)$.

3. Length of the major axis: The length of the major axis is $2a$.

Substitute $a = 20$, the length of the major axis is $2 \times 20 = 40$.

4. Length of the minor axis: The length of the minor axis is $2b$.

Substitute $b = 10$, the length of the minor axis is $2 \times 10 = 20$.

5. Eccentricity: The eccentricity $e$ is given by $e = \frac{c}{a}$.

Substitute $c = 10\sqrt{3}$ and $a = 20$, the eccentricity is $\displaystyle e = \frac{10\sqrt{3}}{20}$.

Simplify the fraction:

$\displaystyle e = \frac{\cancel{10}\sqrt{3}}{\cancel{20}_{2}} = \frac{\sqrt{3}}{2}$

6. Length of the latus rectum: The length of the latus rectum is $\displaystyle \frac{2b^2}{a}$.

Substitute $b^2 = 100$ and $a = 20$, the length of the latus rectum is $\displaystyle \frac{2 \times 100}{20}$.

Simplify the expression:

$\displaystyle \frac{200}{20} = 10$

---

Summary of Results:

Foci: $(0, \pm 10\sqrt{3})$

Vertices: $(0, \pm 20)$

Length of Major Axis: 40

Length of Minor Axis: 20

Eccentricity: $\frac{\sqrt{3}}{2}$

Length of Latus Rectum: 10

Given:

The equation of the ellipse is $36x^2 + 4y^2 = 144$.

---

To Find:

The coordinates of the foci, the vertices, the length of the major axis, the length of the minor axis, the eccentricity, and the length of the latus rectum of the given ellipse.

---

Solution:

The given equation of the ellipse is:

$36x^2 + 4y^2 = 144$

To convert this to the standard form of an ellipse, we need to make the right-hand side equal to 1.

Divide the entire equation by 144:

$\displaystyle \frac{36x^2}{144} + \frac{4y^2}{144} = \frac{144}{144}$

Simplify the fractions:

$\displaystyle \frac{\cancel{36}x^2}{\cancel{144}_{4}} + \frac{\cancel{4}y^2}{\cancel{144}_{36}} = 1$

This equation is in the standard form $\frac{x^{2}}{b^2} + \frac{y^{2}}{a^2} = 1$, where the centre of the ellipse is at the origin $(0, 0)$, because the denominator of the $y^2$ term (36) is greater than the denominator of the $x^2$ term (4).

Comparing equation (i) with the standard form $\frac{x^{2}}{b^2} + \frac{y^{2}}{a^2} = 1$, we have:

$a^2 = 36 \implies a = \sqrt{36} = 6$ (since $a > 0$)

$b^2 = 4 \implies b = \sqrt{4} = 2$ (since $b > 0$)

Since $a^2 = 36 > b^2 = 4$, the major axis is along the y-axis.

Now we can find the required properties of the ellipse with the major axis along the y-axis:

1. Coordinates of the foci: For an ellipse with the major axis along the y-axis, the coordinates of the foci are $(0, \pm c)$, where $c^2 = a^2 - b^2$.

Calculate $c^2$:

$c^2 = 36 - 4$

$c^2 = 32$

$c = \sqrt{32} = \sqrt{16 \times 2} = 4\sqrt{2}$ (since $c > 0$)

The coordinates of the foci are $(0, \pm 4\sqrt{2})$, i.e., $(0, 4\sqrt{2})$ and $(0, -4\sqrt{2})$.

2. Vertices: For an ellipse with the major axis along the y-axis, the coordinates of the vertices are $(0, \pm a)$.

Substitute $a = 6$, the coordinates of the vertices are $(0, \pm 6)$, i.e., $(0, 6)$ and $(0, -6)$.

3. Length of the major axis: The length of the major axis is $2a$.

Substitute $a = 6$, the length of the major axis is $2 \times 6 = 12$.

4. Length of the minor axis: The length of the minor axis is $2b$.

Substitute $b = 2$, the length of the minor axis is $2 \times 2 = 4$.

5. Eccentricity: The eccentricity $e$ is given by $e = \frac{c}{a}$.

Substitute $c = 4\sqrt{2}$ and $a = 6$, the eccentricity is $\displaystyle e = \frac{4\sqrt{2}}{6}$.

Simplify the fraction:

$\displaystyle e = \frac{\cancel{4}^{2}\sqrt{2}}{\cancel{6}_{3}} = \frac{2\sqrt{2}}{3}$

6. Length of the latus rectum: The length of the latus rectum is $\displaystyle \frac{2b^2}{a}$.

Substitute $b^2 = 4$ and $a = 6$, the length of the latus rectum is $\displaystyle \frac{2 \times 4}{6}$.

Simplify the fraction:

$\displaystyle \frac{8}{6} = \frac{\cancel{8}^{4}}{\cancel{6}_{3}} = \frac{4}{3}$

---

Summary of Results:

Foci: $(0, \pm 4\sqrt{2})$

Vertices: $(0, \pm 6)$

Length of Major Axis: 12

Length of Minor Axis: 4

Eccentricity: $\frac{2\sqrt{2}}{3}$

Length of Latus Rectum: $\frac{4}{3}$

Given:

The equation of the ellipse is $16x^2 + y^2 = 16$.

---

To Find:

The coordinates of the foci, the vertices, the length of the major axis, the length of the minor axis, the eccentricity, and the length of the latus rectum of the given ellipse.

---

Solution:

The given equation of the ellipse is:

$16x^2 + y^2 = 16$

To convert this to the standard form of an ellipse, we need to make the right-hand side equal to 1.

Divide the entire equation by 16:

$\displaystyle \frac{16x^2}{16} + \frac{y^2}{16} = \frac{16}{16}$

Simplify the fractions:

$\displaystyle \frac{\cancel{16}x^2}{\cancel{16}_{1}} + \frac{y^2}{16} = 1$

This equation is in the standard form $\frac{x^{2}}{b^2} + \frac{y^{2}}{a^2} = 1$, where the centre of the ellipse is at the origin $(0, 0)$, because the denominator of the $y^2$ term (16) is greater than the denominator of the $x^2$ term (1).

Comparing equation (i) with the standard form $\frac{x^{2}}{b^2} + \frac{y^{2}}{a^2} = 1$, we have:

$a^2 = 16 \implies a = \sqrt{16} = 4$ (since $a > 0$)

$b^2 = 1 \implies b = \sqrt{1} = 1$ (since $b > 0$)

Since $a^2 = 16 > b^2 = 1$, the major axis is along the y-axis.

Now we can find the required properties of the ellipse with the major axis along the y-axis:

1. Coordinates of the foci: For an ellipse with the major axis along the y-axis, the coordinates of the foci are $(0, \pm c)$, where $c^2 = a^2 - b^2$.

Calculate $c^2$:

$c^2 = 16 - 1$

$c^2 = 15$

$c = \sqrt{15}$ (since $c > 0$)

The coordinates of the foci are $(0, \pm \sqrt{15})$, i.e., $(0, \sqrt{15})$ and $(0, -\sqrt{15})$.

2. Vertices: For an ellipse with the major axis along the y-axis, the coordinates of the vertices are $(0, \pm a)$.

Substitute $a = 4$, the coordinates of the vertices are $(0, \pm 4)$, i.e., $(0, 4)$ and $(0, -4)$.

3. Length of the major axis: The length of the major axis is $2a$.

Substitute $a = 4$, the length of the major axis is $2 \times 4 = 8$.

4. Length of the minor axis: The length of the minor axis is $2b$.

Substitute $b = 1$, the length of the minor axis is $2 \times 1 = 2$.

5. Eccentricity: The eccentricity $e$ is given by $e = \frac{c}{a}$.

Substitute $c = \sqrt{15}$ and $a = 4$, the eccentricity is $\displaystyle e = \frac{\sqrt{15}}{4}$.

6. Length of the latus rectum: The length of the latus rectum is $\displaystyle \frac{2b^2}{a}$.

Substitute $b^2 = 1$ and $a = 4$, the length of the latus rectum is $\displaystyle \frac{2 \times 1}{4}$.

Simplify the fraction:

$\displaystyle \frac{2}{4} = \frac{\cancel{2}^{1}}{\cancel{4}_{2}} = \frac{1}{2}$

---

Summary of Results:

Foci: $(0, \pm \sqrt{15})$

Vertices: $(0, \pm 4)$

Length of Major Axis: 8

Length of Minor Axis: 2

Eccentricity: $\frac{\sqrt{15}}{4}$

Length of Latus Rectum: $\frac{1}{2}$

Given:

The equation of the ellipse is $4x^2 + 9y^2 = 36$.

---

To Find:

The coordinates of the foci, the vertices, the length of the major axis, the length of the minor axis, the eccentricity, and the length of the latus rectum of the given ellipse.

---

Solution:

The given equation of the ellipse is:

$4x^2 + 9y^2 = 36$

To convert this to the standard form of an ellipse, we need to make the right-hand side equal to 1.

Divide the entire equation by 36:

$\displaystyle \frac{4x^2}{36} + \frac{9y^2}{36} = \frac{36}{36}$

Simplify the fractions:

$\displaystyle \frac{\cancel{4}x^2}{\cancel{36}_{9}} + \frac{\cancel{9}y^2}{\cancel{36}_{4}} = 1$

This equation is in the standard form $\frac{x^{2}}{a^2} + \frac{y^{2}}{b^2} = 1$, where the centre of the ellipse is at the origin $(0, 0)$, because the denominator of the $x^2$ term (9) is greater than the denominator of the $y^2$ term (4).

Comparing equation (i) with the standard form $\frac{x^{2}}{a^2} + \frac{y^{2}}{b^2} = 1$, we have:

$a^2 = 9 \implies a = \sqrt{9} = 3$ (since $a > 0$)

$b^2 = 4 \implies b = \sqrt{4} = 2$ (since $b > 0$)

Since $a^2 = 9 > b^2 = 4$, the major axis is along the x-axis.

Now we can find the required properties of the ellipse with the major axis along the x-axis:

1. Coordinates of the foci: For an ellipse with the major axis along the x-axis, the coordinates of the foci are $(\pm c, 0)$, where $c^2 = a^2 - b^2$.

Calculate $c^2$:

$c^2 = 9 - 4$

$c^2 = 5$

$c = \sqrt{5}$ (since $c > 0$)

The coordinates of the foci are $(\pm \sqrt{5}, 0)$, i.e., $(\sqrt{5}, 0)$ and $(-\sqrt{5}, 0)$.

2. Vertices: For an ellipse with the major axis along the x-axis, the coordinates of the vertices are $(\pm a, 0)$.

Substitute $a = 3$, the coordinates of the vertices are $(\pm 3, 0)$, i.e., $(3, 0)$ and $(-3, 0)$.

3. Length of the major axis: The length of the major axis is $2a$.

Substitute $a = 3$, the length of the major axis is $2 \times 3 = 6$.

4. Length of the minor axis: The length of the minor axis is $2b$.

Substitute $b = 2$, the length of the minor axis is $2 \times 2 = 4$.

5. Eccentricity: The eccentricity $e$ is given by $e = \frac{c}{a}$.

Substitute $c = \sqrt{5}$ and $a = 3$, the eccentricity is $\displaystyle e = \frac{\sqrt{5}}{3}$.

6. Length of the latus rectum: The length of the latus rectum is $\displaystyle \frac{2b^2}{a}$.

Substitute $b^2 = 4$ and $a = 3$, the length of the latus rectum is $\displaystyle \frac{2 \times 4}{3} = \frac{8}{3}$.

---

Summary of Results:

Foci: $(\pm \sqrt{5}, 0)$

Vertices: $(\pm 3, 0)$

Length of Major Axis: 6

Length of Minor Axis: 4

Eccentricity: $\frac{\sqrt{5}}{3}$

Length of Latus Rectum: $\frac{8}{3}$

Given:

The vertices of the ellipse are $(\pm 5, 0)$.

The foci of the ellipse are $(\pm 4, 0)$.

---

To Find:

The equation of the ellipse.

---

Solution:

The given vertices are $(\pm 5, 0)$ and the foci are $(\pm 4, 0)$.

Since the vertices and foci lie on the x-axis and are symmetric with respect to the origin, the centre of the ellipse is at the origin $(0, 0)$, and the major axis is along the x-axis.

For an ellipse with the major axis along the x-axis and centre at the origin, the standard form of the equation is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.

The coordinates of the vertices are $(\pm a, 0)$. Comparing with the given vertices $(\pm 5, 0)$, we have:

So, $a^2 = 5^2 = 25$.

The coordinates of the foci are $(\pm c, 0)$. Comparing with the given foci $(\pm 4, 0)$, we have:

For an ellipse, the relationship between $a$, $b$, and $c$ is $c^2 = a^2 - b^2$. We need to find $b^2$.

Rearrange the formula to solve for $b^2$:

$b^2 = a^2 - c^2$

Substitute the values of $a^2 = 25$ and $c = 4$ (so $c^2 = 4^2 = 16$):

$b^2 = 25 - 16$

Now, substitute the values of $a^2 = 25$ and $b^2 = 9$ into the standard equation of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$:

This is the required equation of the ellipse.

---

Final Equation:

The equation of the ellipse is $\frac{x^2}{25} + \frac{y^2}{9} = 1$.

Given:

The vertices of the ellipse are $(0, \pm 13)$.

The foci of the ellipse are $(0, \pm 5)$.

---

To Find:

The equation of the ellipse.

---

Solution:

The given vertices are $(0, \pm 13)$ and the foci are $(0, \pm 5)$.

Since the vertices and foci lie on the y-axis and are symmetric with respect to the origin, the centre of the ellipse is at the origin $(0, 0)$, and the major axis is along the y-axis.

For an ellipse with the major axis along the y-axis and centre at the origin, the standard form of the equation is $\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$.

The coordinates of the vertices are $(0, \pm a)$. Comparing with the given vertices $(0, \pm 13)$, we have:

So, $a^2 = 13^2 = 169$.

The coordinates of the foci are $(0, \pm c)$. Comparing with the given foci $(0, \pm 5)$, we have:

For an ellipse, the relationship between $a$, $b$, and $c$ is $c^2 = a^2 - b^2$. We need to find $b^2$.

Rearrange the formula to solve for $b^2$:

$b^2 = a^2 - c^2$

Substitute the values of $a^2 = 169$ and $c = 5$ (so $c^2 = 5^2 = 25$):

$b^2 = 169 - 25$

Now, substitute the values of $a^2 = 169$ and $b^2 = 144$ into the standard equation of the ellipse $\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$:

This is the required equation of the ellipse.

---

Final Equation:

The equation of the ellipse is $\frac{x^2}{144} + \frac{y^2}{169} = 1$.

Given:

The vertices of the ellipse are $(\pm 6, 0)$.

The foci of the ellipse are $(\pm 4, 0)$.

---

To Find:

The equation of the ellipse.

---

Solution:

The given vertices are $(\pm 6, 0)$ and the foci are $(\pm 4, 0)$.

Since the vertices and foci lie on the x-axis and are symmetric with respect to the origin, the centre of the ellipse is at the origin $(0, 0)$, and the major axis is along the x-axis.

For an ellipse with the major axis along the x-axis and centre at the origin, the standard form of the equation is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.

The coordinates of the vertices are $(\pm a, 0)$. Comparing with the given vertices $(\pm 6, 0)$, we have:

So, $a^2 = 6^2 = 36$.

The coordinates of the foci are $(\pm c, 0)$. Comparing with the given foci $(\pm 4, 0)$, we have:

For an ellipse, the relationship between $a$, $b$, and $c$ is $c^2 = a^2 - b^2$. We need to find $b^2$.

Rearrange the formula to solve for $b^2$:

$b^2 = a^2 - c^2$

Substitute the values of $a^2 = 36$ and $c = 4$ (so $c^2 = 4^2 = 16$):

$b^2 = 36 - 16$

Now, substitute the values of $a^2 = 36$ and $b^2 = 20$ into the standard equation of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$:

This is the required equation of the ellipse.

---

Final Equation:

The equation of the ellipse is $\frac{x^2}{36} + \frac{y^2}{20} = 1$.

Given:

The ends of the major axis are $(\pm 3, 0)$.

The ends of the minor axis are $(0, \pm 2)$.

---

To Find:

The equation of the ellipse.

---

Solution:

The ends of the major axis are $(\pm 3, 0)$. These points lie on the x-axis, and they are symmetric about the origin. This indicates that the centre of the ellipse is at the origin $(0, 0)$, and the major axis lies along the x-axis.

For an ellipse with its centre at the origin and major axis along the x-axis, the standard form of the equation is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, where $a > b$.

The coordinates of the ends of the major axis are $(\pm a, 0)$. Comparing with the given ends of the major axis $(\pm 3, 0)$, we have:

So, $a^2 = 3^2 = 9$.

The coordinates of the ends of the minor axis are $(0, \pm b)$. Comparing with the given ends of the minor axis $(0, \pm 2)$, we have:

So, $b^2 = 2^2 = 4$.

We can verify that $a = 3 > b = 2$, which is consistent with the major axis being along the x-axis.

Now, substitute the values of $a^2 = 9$ and $b^2 = 4$ into the standard equation of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$:

This is the required equation of the ellipse.

---

Final Equation:

The equation of the ellipse is $\frac{x^2}{9} + \frac{y^2}{4} = 1$.

Given:

The ends of the major axis are $(0, \pm \sqrt{5})$.

The ends of the minor axis are $(\pm 1, 0)$.

---

To Find:

The equation of the ellipse.

---

Solution:

The ends of the major axis are $(0, \pm \sqrt{5})$. These points lie on the y-axis, and they are symmetric about the origin. This indicates that the centre of the ellipse is at the origin $(0, 0)$, and the major axis lies along the y-axis.

For an ellipse with its centre at the origin and major axis along the y-axis, the standard form of the equation is $\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$, where $a > b$.

The coordinates of the ends of the major axis are $(0, \pm a)$. Comparing with the given ends of the major axis $(0, \pm \sqrt{5})$, we have:

So, $a^2 = (\sqrt{5})^2 = 5$.

The coordinates of the ends of the minor axis are $(\pm b, 0)$. Comparing with the given ends of the minor axis $(\pm 1, 0)$, we have:

So, $b^2 = 1^2 = 1$.

We can verify that $a = \sqrt{5} > b = 1$ (since $\sqrt{5} \approx 2.236$), which is consistent with the major axis being along the y-axis.

Now, substitute the values of $a^2 = 5$ and $b^2 = 1$ into the standard equation of the ellipse $\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$:

This is the required equation of the ellipse.

---

Final Equation:

The equation of the ellipse is $x^2 + \frac{y^2}{5} = 1$.

Alternatively, multiply by 5 to clear the denominator: $5x^2 + y^2 = 5$ or $5x^2 + y^2 - 5 = 0$.

Given:

The length of the major axis of the ellipse is 26.

The foci of the ellipse are $(\pm 5, 0)$.

---

To Find:

The equation of the ellipse.

---

Solution:

The length of the major axis is given as 26. The length of the major axis is equal to $2a$.

$\displaystyle 2a = 26$

Divide by 2:

So, $a^2 = 13^2 = 169$.

The foci are given as $(\pm 5, 0)$. The coordinates of the foci for an ellipse centred at the origin are $(\pm c, 0)$ or $(0, \pm c)$.

Since the foci are $(\pm 5, 0)$, the centre of the ellipse is at the origin $(0, 0)$, and the major axis is along the x-axis.

The coordinates of the foci are $(\pm c, 0)$. Comparing with the given foci $(\pm 5, 0)$, we have:

We know $a = 13$ (so $a^2 = 169$) and $c = 5$ (so $c^2 = 25$). We need to find $b^2$.

The relationship between $a$, $b$, and $c$ for an ellipse is $c^2 = a^2 - b^2$.

Rearrange the formula to solve for $b^2$:

$b^2 = a^2 - c^2$

Substitute the values of $a^2 = 169$ and $c^2 = 25$:

$b^2 = 169 - 25$

For an ellipse with the major axis along the x-axis and centre at the origin, the standard form of the equation is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.

Substitute the values of $a^2 = 169$ and $b^2 = 144$ into the standard equation:

This is the required equation of the ellipse.

---

Final Equation:

The equation of the ellipse is $\frac{x^2}{169} + \frac{y^2}{144} = 1$.

Given:

The length of the minor axis of the ellipse is 16.

The foci of the ellipse are $(0, \pm 6)$.

---

To Find:

The equation of the ellipse.

---

Solution:

The length of the minor axis is given as 16. The length of the minor axis is equal to $2b$.

$\displaystyle 2b = 16$

Divide by 2:

So, $b^2 = 8^2 = 64$.

The foci are given as $(0, \pm 6)$. The coordinates of the foci for an ellipse centred at the origin are $(\pm c, 0)$ or $(0, \pm c)$.

Since the foci are $(0, \pm 6)$, the centre of the ellipse is at the origin $(0, 0)$, and the major axis is along the y-axis.

The coordinates of the foci are $(0, \pm c)$. Comparing with the given foci $(0, \pm 6)$, we have:

So, $c^2 = 6^2 = 36$.

For an ellipse, the relationship between $a$, $b$, and $c$ is $c^2 = a^2 - b^2$. We need to find $a^2$.

Rearrange the formula to solve for $a^2$:

$a^2 = b^2 + c^2$

Substitute the values of $b^2 = 64$ and $c^2 = 36$:

$a^2 = 64 + 36$

For an ellipse with the major axis along the y-axis and centre at the origin, the standard form of the equation is $\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$.

Substitute the values of $a^2 = 100$ and $b^2 = 64$ into the standard equation:

This is the required equation of the ellipse.

---

Final Equation:

The equation of the ellipse is $\frac{x^2}{64} + \frac{y^2}{100} = 1$.

Given:

The foci of the ellipse are $(\pm 3, 0)$.

The value of $a$ is 4.

---

To Find:

The equation of the ellipse.

---

Solution:

The foci are given as $(\pm 3, 0)$. The coordinates of the foci for an ellipse centred at the origin are $(\pm c, 0)$ or $(0, \pm c)$.

Since the foci are $(\pm 3, 0)$, the centre of the ellipse is at the origin $(0, 0)$, and the major axis is along the x-axis.

The coordinates of the foci are $(\pm c, 0)$. Comparing with the given foci $(\pm 3, 0)$, we have:

So, $c^2 = 3^2 = 9$.

We are given that $a = 4$.

So, $a^2 = 4^2 = 16$.

For an ellipse, the relationship between $a$, $b$, and $c$ is $c^2 = a^2 - b^2$. We need to find $b^2$.

Rearrange the formula to solve for $b^2$:

$b^2 = a^2 - c^2$

Substitute the values of $a^2 = 16$ and $c^2 = 9$:

$b^2 = 16 - 9$

For an ellipse with the major axis along the x-axis and centre at the origin, the standard form of the equation is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.

Substitute the values of $a^2 = 16$ and $b^2 = 7$ into the standard equation:

This is the required equation of the ellipse.

---

Final Equation:

The equation of the ellipse is $\frac{x^2}{16} + \frac{y^2}{7} = 1$.

Given:

The value of $b$ is 3.

The value of $c$ is 4.

The centre of the ellipse is at the origin $(0, 0)$.

The foci are on the x-axis.

---

To Find:

The equation of the ellipse.

---

Solution:

The centre of the ellipse is at the origin $(0, 0)$ and the foci are on the x-axis. This means the major axis is along the x-axis.

For an ellipse with the major axis along the x-axis and centre at the origin, the standard form of the equation is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, where $a > b$ and the foci are at $(\pm c, 0)$.

We are given $b = 3$, so $b^2 = 3^2 = 9$.

We are given $c = 4$, so $c^2 = 4^2 = 16$.

For an ellipse, the relationship between $a$, $b$, and $c$ is $c^2 = a^2 - b^2$. We need to find $a^2$.

Rearrange the formula to solve for $a^2$:

$a^2 = b^2 + c^2$

Substitute the values of $b^2 = 9$ and $c^2 = 16$:

$a^2 = 9 + 16$

So, $a = \sqrt{25} = 5$. We can check that $a = 5 > b = 3$, which is consistent with the major axis being along the x-axis.

Now, substitute the values of $a^2 = 25$ and $b^2 = 9$ into the standard equation of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$:

This is the required equation of the ellipse.

---

Final Equation:

The equation of the ellipse is $\frac{x^2}{25} + \frac{y^2}{9} = 1$.

Given:

The centre of the ellipse is at $(0, 0)$.

The major axis of the ellipse is on the y-axis.

The ellipse passes through the points $(3, 2)$ and $(1, 6)$.

---

To Find:

The equation of the ellipse.

---

Solution:

Since the centre is at the origin $(0, 0)$ and the major axis is on the y-axis, the standard form of the equation of the ellipse is $\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$, where $a > b$.

Since the ellipse passes through the point $(3, 2)$, substitute $x = 3$ and $y = 2$ into the equation:

Since the ellipse also passes through the point $(1, 6)$, substitute $x = 1$ and $y = 6$ into the equation:

Now we have a system of two linear equations in terms of $\frac{1}{a^2}$ and $\frac{1}{b^2}$. Let $A = \frac{1}{a^2}$ and $B = \frac{1}{b^2}$. The system becomes:

1) $4A + 9B = 1$

2) $36A + 1B = 1$

From equation (2), we can express $B$ in terms of $A$:

Substitute equation (3) into equation (1):

$4A + 9(1 - 36A) = 1$

$4A + 9 - 324A = 1$

Combine like terms:

$-320A + 9 = 1$

$-320A = 1 - 9$

$-320A = -8$

$\displaystyle A = \frac{-8}{-320} = \frac{8}{320}$

Simplify the fraction by dividing both numerator and denominator by 8:

$\displaystyle A = \frac{\cancel{8}^{1}}{\cancel{320}_{40}} = \frac{1}{40}$

So, $\displaystyle \frac{1}{a^2} = \frac{1}{40}$. This gives $a^2 = 40$.

Now substitute the value of $A = \frac{1}{40}$ back into equation (3) to find $B$:

$\displaystyle B = 1 - 36\left(\frac{1}{40}\right)$

$\displaystyle B = 1 - \frac{36}{40}$

Simplify the fraction $\frac{36}{40}$ by dividing by 4:

$\displaystyle \frac{36}{40} = \frac{\cancel{36}^{9}}{\cancel{40}_{10}} = \frac{9}{10}$

$\displaystyle B = 1 - \frac{9}{10}$

$\displaystyle B = \frac{10 - 9}{10} = \frac{1}{10}$

So, $\displaystyle \frac{1}{b^2} = \frac{1}{10}$. This gives $b^2 = 10$.

We must have $a^2 > b^2$ since the major axis is on the y-axis. We found $a^2 = 40$ and $b^2 = 10$. Since $40 > 10$, our values are consistent.

Now, substitute the values of $a^2 = 40$ and $b^2 = 10$ into the standard equation of the ellipse $\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$:

This is the required equation of the ellipse.

---

Final Equation:

The equation of the ellipse is $\frac{x^2}{10} + \frac{y^2}{40} = 1$.

Given:

The major axis of the ellipse is on the x-axis.

The ellipse passes through the points $(4, 3)$ and $(6, 2)$.

---

To Find:

The equation of the ellipse.

---

Solution:

Since the major axis is on the x-axis, and assuming the centre of the ellipse is at the origin $(0, 0)$, the standard form of the equation is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, where $a > b$.

Since the ellipse passes through the point $(4, 3)$, substitute $x = 4$ and $y = 3$ into the equation:

Since the ellipse also passes through the point $(6, 2)$, substitute $x = 6$ and $y = 2$ into the equation:

Now we have a system of two linear equations in terms of $\frac{1}{a^2}$ and $\frac{1}{b^2}$. Let $A = \frac{1}{a^2}$ and $B = \frac{1}{b^2}$. The system becomes:

1) $16A + 9B = 1$

2) $36A + 4B = 1$

We can solve this system using elimination or substitution.

Multiply equation (1) by 4 and equation (2) by 9 to eliminate $B$:

$4 \times (16A + 9B = 1) \implies 64A + 36B = 4$

$9 \times (36A + 4B = 1) \implies 324A + 36B = 9$

Subtract equation (3) from equation (4):

$(324A + 36B) - (64A + 36B) = 9 - 4$

$324A - 64A + 36B - 36B = 5$

$260A = 5$

$\displaystyle A = \frac{5}{260}$

Simplify the fraction by dividing both numerator and denominator by 5:

$\displaystyle A = \frac{\cancel{5}^{1}}{\cancel{260}_{52}} = \frac{1}{52}$

So, $\displaystyle \frac{1}{a^2} = \frac{1}{52}$. This gives $a^2 = 52$.

Now substitute the value of $A = \frac{1}{52}$ back into equation (1):

$\displaystyle 16\left(\frac{1}{52}\right) + 9B = 1$

$\displaystyle \frac{16}{52} + 9B = 1$

Simplify the fraction $\frac{16}{52}$ by dividing by 4:

$\displaystyle \frac{16}{52} = \frac{\cancel{16}^{4}}{\cancel{52}_{13}} = \frac{4}{13}$

$\displaystyle \frac{4}{13} + 9B = 1$

$9B = 1 - \frac{4}{13}$

$\displaystyle 9B = \frac{13 - 4}{13} = \frac{9}{13}$

Divide both sides by 9:

$\displaystyle B = \frac{9}{13 \times 9} = \frac{\cancel{9}}{\cancel{9} \times 13} = \frac{1}{13}$

So, $\displaystyle \frac{1}{b^2} = \frac{1}{13}$. This gives $b^2 = 13$.

We must have $a^2 > b^2$ since the major axis is on the x-axis. We found $a^2 = 52$ and $b^2 = 13$. Since $52 > 13$, our values are consistent.

Now, substitute the values of $a^2 = 52$ and $b^2 = 13$ into the standard equation of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$:

This is the required equation of the ellipse.

---

Final Equation:

The equation of the ellipse is $\frac{x^2}{52} + \frac{y^2}{13} = 1$.

Given:

Two hyperbola equations: (i) $\frac{x^{2}}{9} - \frac{y^{2}}{16} = 1$ and (ii) $y^2 – 16x^2 = 16$.

---

To Find:

For each hyperbola, find the coordinates of the foci and vertices, the eccentricity, and the length of the latus rectum.

---

Solution for (i) $\frac{x^{2}}{9} - \frac{y^{2}}{16} = 1$

The equation is in the standard form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.

Since the $x^2$ term is positive, the transverse axis is along the x-axis.

Comparing the given equation with the standard form, we have:

$a^2 = 9 \implies a = 3$

$b^2 = 16 \implies b = 4$

For a hyperbola, the relationship between $a, b,$ and $c$ (distance from centre to focus) is $c^2 = a^2 + b^2$.

$c^2 = 9 + 16 = 25$

$\implies c = \sqrt{25} = 5$

Now, we find the required properties:

1. Foci: The coordinates are $(\pm c, 0)$. Thus, the foci are $(\pm 5, 0)$.

2. Vertices: The coordinates are $(\pm a, 0)$. Thus, the vertices are $(\pm 3, 0)$.

3. Eccentricity ($e$): The formula is $e = \frac{c}{a}$.

$e = \frac{5}{3}$ (Note: For a hyperbola, $e > 1$).

4. Latus Rectum: The length is given by $\frac{2b^2}{a}$.

Length = $\frac{2(16)}{3} = \frac{32}{3}$.

---

Solution for (ii) $y^2 – 16x^2 = 16$

First, we convert the equation to standard form by dividing both sides by 16:

$\frac{y^2}{16} - \frac{16x^2}{16} = \frac{16}{16}$

This equation is in the standard form $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$.

Since the $y^2$ term is positive, the transverse axis is along the y-axis.

Comparing with the standard form, we have:

$a^2 = 16 \implies a = 4$

$b^2 = 1 \implies b = 1$

Now, we find $c$ using $c^2 = a^2 + b^2$.

$c^2 = 16 + 1 = 17$

$\implies c = \sqrt{17}$

Now, we find the required properties:

1. Foci: The coordinates are $(0, \pm c)$. Thus, the foci are $(0, \pm \sqrt{17})$.

2. Vertices: The coordinates are $(0, \pm a)$. Thus, the vertices are $(0, \pm 4)$.

3. Eccentricity ($e$): The formula is $e = \frac{c}{a}$.

$e = \frac{\sqrt{17}}{4}$

4. Latus Rectum: The length is given by $\frac{2b^2}{a}$.

Length = $\frac{2(1)}{4} = \frac{1}{2}$.

Given:

The foci of the hyperbola are $(0, \pm 3)$.

The vertices are $\left( 0, \pm \frac{\sqrt{11}}{2}\right)$.

---

To Find:

The equation of the hyperbola.

---

Solution:

Since the foci and vertices are on the y-axis, the transverse axis is along the y-axis and the centre is at the origin $(0,0)$.

The standard equation for this type of hyperbola is:

The coordinates of the vertices are $(0, \pm a)$. From the given vertices, we have:

$a = \frac{\sqrt{11}}{2} \implies a^2 = \left(\frac{\sqrt{11}}{2}\right)^2 = \frac{11}{4}$

The coordinates of the foci are $(0, \pm c)$. From the given foci, we have:

$c = 3 \implies c^2 = 9$

For a hyperbola, the relationship between $a, b,$ and $c$ is $c^2 = a^2 + b^2$. We can find $b^2$ as follows:

$b^2 = c^2 - a^2$

$b^2 = 9 - \frac{11}{4} = \frac{36 - 11}{4} = \frac{25}{4}$

Now, substitute the values of $a^2 = \frac{11}{4}$ and $b^2 = \frac{25}{4}$ into the standard equation (i):

$\frac{y^2}{(11/4)} - \frac{x^2}{(25/4)} = 1$

$\implies \frac{4y^2}{11} - \frac{4x^2}{25} = 1$

This is the required equation of the hyperbola.

---

Final Answer:

The equation of the hyperbola is $\frac{4y^2}{11} - \frac{4x^2}{25} = 1$.

Given:

The foci of the hyperbola are $(0, \pm 12)$.

The length of the latus rectum is 36.

---

To Find:

The equation of the hyperbola.

---

Solution:

Since the foci are at $(0, \pm 12)$, they lie on the y-axis. This means the transverse axis is along the y-axis and the centre is at the origin $(0,0)$.

The standard equation for this type of hyperbola is:

From the foci $(0, \pm 12)$, we know that $c = 12$, so $c^2 = 144$.

The length of the latus rectum is given by the formula $\frac{2b^2}{a}$. We are given this length is 36.

$\frac{2b^2}{a} = 36$

$\implies 2b^2 = 36a$

The relationship for a hyperbola is $c^2 = a^2 + b^2$. Substituting $c^2 = 144$ and $b^2 = 18a$:

$144 = a^2 + 18a$

Rearranging this into a quadratic equation:

$a^2 + 18a - 144 = 0$

We solve this by factoring. We need two numbers that multiply to -144 and add to 18. These numbers are 24 and -6.

$(a + 24)(a - 6) = 0$

This gives two possible solutions for $a$: $a = -24$ or $a = 6$.

Since $a$ represents a distance, it must be positive. Therefore, $a=6$.

So, $a^2 = 36$.

Now, we find $b^2$ using equation (ii):

$b^2 = 18a = 18(6) = 108$

We have $a^2 = 36$ and $b^2 = 108$. Substituting these values into the standard equation (i):

$\frac{y^2}{36} - \frac{x^2}{108} = 1$

---

Final Answer:

The required equation of the hyperbola is $\frac{y^2}{36} - \frac{x^2}{108} = 1$.

Given:

The equation of the hyperbola is $\frac{x^{2}}{16}$ - $\frac{y^{2}}{9}$ = 1.

---

To Find:

The coordinates of the foci, the vertices, the eccentricity, and the length of the latus rectum of the given hyperbola.

---

Solution:

The given equation of the hyperbola is:

This equation is in the standard form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, where the centre is at the origin $(0, 0)$ and the transverse axis is along the x-axis (since the $x^2$ term is positive).

Comparing equation (i) with the standard form, we have:

$a^2 = 16 \implies a = \sqrt{16} = 4$ (since $a > 0$)

$b^2 = 9 \implies b = \sqrt{9} = 3$ (since $b > 0$)

Now we find the value of $c$ using the relation $c^2 = a^2 + b^2$ for a hyperbola:

$c^2 = 16 + 9$

$c^2 = 25$

$c = \sqrt{25} = 5$ (since $c > 0$)

Using the values of $a, b, c$ and the fact that the transverse axis is along the x-axis:

1. Coordinates of the foci: The foci are at $(\pm c, 0)$.

Substituting $c = 5$, the coordinates of the foci are $(\pm 5, 0)$, i.e., $(5, 0)$ and $(-5, 0)$.

2. Vertices: The vertices are at $(\pm a, 0)$.

Substituting $a = 4$, the coordinates of the vertices are $(\pm 4, 0)$, i.e., $(4, 0)$ and $(-4, 0)$.

3. Eccentricity: The eccentricity $e$ is given by $e = \frac{c}{a}$.

Substitute $c = 5$ and $a = 4$, the eccentricity is $\displaystyle e = \frac{5}{4}$.

4. Length of the latus rectum: The length of the latus rectum is $\displaystyle \frac{2b^2}{a}$.

Substitute $b^2 = 9$ and $a = 4$, the length of the latus rectum is $\displaystyle \frac{2 \times 9}{4}$.

Simplify the fraction:

$\displaystyle \frac{18}{4} = \frac{\cancel{18}^{9}}{\cancel{4}_{2}} = \frac{9}{2}$

---

Summary of Results:

Foci: $(\pm 5, 0)$

Vertices: $(\pm 4, 0)$

Eccentricity: $\frac{5}{4}$

Length of Latus Rectum: $\frac{9}{2}$

Given:

The equation of the hyperbola is $\frac{y^{2}}{9}$ - $\frac{x^{2}}{27}$ = 1.

---

To Find:

The coordinates of the foci, the vertices, the eccentricity, and the length of the latus rectum of the given hyperbola.

---

Solution:

The given equation of the hyperbola is:

This equation is in the standard form $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$, where the centre is at the origin $(0, 0)$ and the transverse axis is along the y-axis (since the $y^2$ term is positive).

Comparing equation (i) with the standard form, we have:

$a^2 = 9 \implies a = \sqrt{9} = 3$ (since $a > 0$)

$b^2 = 27 \implies b = \sqrt{27} = \sqrt{9 \times 3} = 3\sqrt{3}$ (since $b > 0$)

Now we find the value of $c$ using the relation $c^2 = a^2 + b^2$ for a hyperbola:

$c^2 = 9 + 27$

$c^2 = 36$

$c = \sqrt{36} = 6$ (since $c > 0$)

Using the values of $a, b, c$ and the fact that the transverse axis is along the y-axis:

1. Coordinates of the foci: The foci are at $(0, \pm c)$.

Substituting $c = 6$, the coordinates of the foci are $(0, \pm 6)$, i.e., $(0, 6)$ and $(0, -6)$.

2. Vertices: The vertices are at $(0, \pm a)$.

Substituting $a = 3$, the coordinates of the vertices are $(0, \pm 3)$, i.e., $(0, 3)$ and $(0, -3)$.

3. Eccentricity: The eccentricity $e$ is given by $e = \frac{c}{a}$.

Substitute $c = 6$ and $a = 3$, the eccentricity is $\displaystyle e = \frac{6}{3} = 2$.

4. Length of the latus rectum: The length of the latus rectum is $\displaystyle \frac{2b^2}{a}$.

Substitute $b^2 = 27$ and $a = 3$, the length of the latus rectum is $\displaystyle \frac{2 \times 27}{3}$.

Simplify the expression:

$\displaystyle \frac{54}{3} = 18$

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Summary of Results:

Foci: $(0, \pm 6)$

Vertices: $(0, \pm 3)$

Eccentricity: 2

Length of Latus Rectum: 18

Given:

The equation of the hyperbola is $9y^2 – 4x^2 = 36$.

---

To Find:

The coordinates of the foci, the vertices, the eccentricity, and the length of the latus rectum of the given hyperbola.

---

Solution:

The given equation of the hyperbola is:

$9y^2 – 4x^2 = 36$

To convert this to the standard form, divide the entire equation by 36:

$\displaystyle \frac{9y^2}{36} - \frac{4x^2}{36} = \frac{36}{36}$

Simplify the fractions:

$\displaystyle \frac{\cancel{9}y^2}{\cancel{36}_{4}} - \frac{\cancel{4}x^2}{\cancel{36}_{9}} = 1$

This equation is in the standard form $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$, where the centre is at the origin $(0, 0)$ and the transverse axis is along the y-axis (since the $y^2$ term is positive).

Comparing equation (i) with the standard form, we have:

$a^2 = 4 \implies a = \sqrt{4} = 2$ (since $a > 0$)

$b^2 = 9 \implies b = \sqrt{9} = 3$ (since $b > 0$)

Now we find the value of $c$ using the relation $c^2 = a^2 + b^2$ for a hyperbola:

$c^2 = 4 + 9$

$c^2 = 13$

$c = \sqrt{13}$ (since $c > 0$)

Using the values of $a, b, c$ and the fact that the transverse axis is along the y-axis:

1. Coordinates of the foci: The foci are at $(0, \pm c)$.

Substituting $c = \sqrt{13}$, the coordinates of the foci are $(0, \pm \sqrt{13})$, i.e., $(0, \sqrt{13})$ and $(0, -\sqrt{13})$.

2. Vertices: The vertices are at $(0, \pm a)$.

Substituting $a = 2$, the coordinates of the vertices are $(0, \pm 2)$, i.e., $(0, 2)$ and $(0, -2)$.

3. Eccentricity: The eccentricity $e$ is given by $e = \frac{c}{a}$.

Substitute $c = \sqrt{13}$ and $a = 2$, the eccentricity is $\displaystyle e = \frac{\sqrt{13}}{2}$.

4. Length of the latus rectum: The length of the latus rectum is $\displaystyle \frac{2b^2}{a}$.

Substitute $b^2 = 9$ and $a = 2$, the length of the latus rectum is $\displaystyle \frac{2 \times 9}{2}$.

Simplify the expression:

$\displaystyle \frac{18}{2} = 9$

---

Summary of Results:

Foci: $(0, \pm \sqrt{13})$

Vertices: $(0, \pm 2)$

Eccentricity: $\frac{\sqrt{13}}{2}$

Length of Latus Rectum: 9

Given:

The equation of the hyperbola is $16x^2 – 9y^2 = 576$.

---

To Find:

The coordinates of the foci, the vertices, the eccentricity, and the length of the latus rectum of the given hyperbola.

---

Solution:

The given equation of the hyperbola is:

$16x^2 – 9y^2 = 576$

To convert this to the standard form, divide the entire equation by 576:

$\displaystyle \frac{16x^2}{576} - \frac{9y^2}{576} = \frac{576}{576}$

Simplify the fractions:

$\displaystyle \frac{\cancel{16}x^2}{\cancel{576}_{36}} - \frac{\cancel{9}y^2}{\cancel{576}_{64}} = 1$

This equation is in the standard form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, where the centre is at the origin $(0, 0)$ and the transverse axis is along the x-axis (since the $x^2$ term is positive).

Comparing equation (i) with the standard form, we have:

$a^2 = 36 \implies a = \sqrt{36} = 6$ (since $a > 0$)

$b^2 = 64 \implies b = \sqrt{64} = 8$ (since $b > 0$)

Now we find the value of $c$ using the relation $c^2 = a^2 + b^2$ for a hyperbola:

$c^2 = 36 + 64$

$c^2 = 100$

$c = \sqrt{100} = 10$ (since $c > 0$)

Using the values of $a, b, c$ and the fact that the transverse axis is along the x-axis:

1. Coordinates of the foci: The foci are at $(\pm c, 0)$.

Substituting $c = 10$, the coordinates of the foci are $(\pm 10, 0)$, i.e., $(10, 0)$ and $(-10, 0)$.

2. Vertices: The vertices are at $(\pm a, 0)$.

Substituting $a = 6$, the coordinates of the vertices are $(\pm 6, 0)$, i.e., $(6, 0)$ and $(-6, 0)$.

3. Eccentricity: The eccentricity $e$ is given by $e = \frac{c}{a}$.

Substitute $c = 10$ and $a = 6$, the eccentricity is $\displaystyle e = \frac{10}{6}$.

Simplify the fraction:

$\displaystyle e = \frac{\cancel{10}^{5}}{\cancel{6}_{3}} = \frac{5}{3}$

4. Length of the latus rectum: The length of the latus rectum is $\displaystyle \frac{2b^2}{a}$.

Substitute $b^2 = 64$ and $a = 6$, the length of the latus rectum is $\displaystyle \frac{2 \times 64}{6}$.

Simplify the fraction:

$\displaystyle \frac{128}{6} = \frac{\cancel{128}^{64}}{\cancel{6}_{3}} = \frac{64}{3}$

---

Summary of Results:

Foci: $(\pm 10, 0)$

Vertices: $(\pm 6, 0)$

Eccentricity: $\frac{5}{3}$

Length of Latus Rectum: $\frac{64}{3}$

Given:

The equation of the hyperbola is $5y^2 – 9x^2 = 36$.

---

To Find:

The coordinates of the foci, the vertices, the eccentricity, and the length of the latus rectum of the given hyperbola.

---

Solution:

The given equation of the hyperbola is:

$5y^2 – 9x^2 = 36$

To convert this to the standard form, divide the entire equation by 36:

$\displaystyle \frac{5y^2}{36} - \frac{9x^2}{36} = \frac{36}{36}$

Rewrite the terms so that the numerators have coefficients of 1:

$\displaystyle \frac{y^2}{36/5} - \frac{x^2}{36/9} = 1$

Simplify the denominators:

This equation is in the standard form $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$, where the centre is at the origin $(0, 0)$ and the transverse axis is along the y-axis (since the $y^2$ term is positive).

Comparing equation (i) with the standard form, we have:

$a^2 = \frac{36}{5} \implies a = \sqrt{\frac{36}{5}} = \frac{6}{\sqrt{5}}$ (since $a > 0$)

$b^2 = 4 \implies b = \sqrt{4} = 2$ (since $b > 0$)

Now we find the value of $c$ using the relation $c^2 = a^2 + b^2$ for a hyperbola:

$\displaystyle c^2 = \frac{36}{5} + 4$

Find a common denominator:

$\displaystyle c^2 = \frac{36 + 4 \times 5}{5} = \frac{36 + 20}{5} = \frac{56}{5}$

$\displaystyle c = \sqrt{\frac{56}{5}} = \frac{\sqrt{56}}{\sqrt{5}} = \frac{\sqrt{4 \times 14}}{\sqrt{5}} = \frac{2\sqrt{14}}{\sqrt{5}}$ (since $c > 0$)

We can rationalize the denominator: $\displaystyle c = \frac{2\sqrt{14}}{\sqrt{5}} \times \frac{\sqrt{5}}{\sqrt{5}} = \frac{2\sqrt{70}}{5}$.

Using the values of $a, b^2, c$ and the fact that the transverse axis is along the y-axis:

1. Coordinates of the foci: The foci are at $(0, \pm c)$.

Substituting $c = \frac{2\sqrt{70}}{5}$, the coordinates of the foci are $\left(0, \pm \frac{2\sqrt{70}}{5}\right)$.

2. Vertices: The vertices are at $(0, \pm a)$.

Substituting $a = \frac{6}{\sqrt{5}} = \frac{6\sqrt{5}}{5}$, the coordinates of the vertices are $\left(0, \pm \frac{6\sqrt{5}}{5}\right)$.

3. Eccentricity: The eccentricity $e$ is given by $e = \frac{c}{a}$.

Substitute $c = \frac{2\sqrt{70}}{5}$ and $a = \frac{6\sqrt{5}}{5}$:

$\displaystyle e = \frac{2\sqrt{70}/5}{6\sqrt{5}/5} = \frac{2\sqrt{70}}{5} \times \frac{5}{6\sqrt{5}}$

$\displaystyle e = \frac{2\sqrt{70}}{6\sqrt{5}} = \frac{\sqrt{70}}{3\sqrt{5}} = \frac{\sqrt{14 \times 5}}{3\sqrt{5}} = \frac{\sqrt{14} \times \sqrt{5}}{3\sqrt{5}} = \frac{\sqrt{14}}{3}$

Alternatively, using $c = \sqrt{56/5}$ and $a = \sqrt{36/5}$: $\displaystyle e = \frac{\sqrt{56/5}}{\sqrt{36/5}} = \sqrt{\frac{56/5}{36/5}} = \sqrt{\frac{56}{36}} = \sqrt{\frac{14}{9}} = \frac{\sqrt{14}}{3}$.

4. Length of the latus rectum: The length of the latus rectum is $\displaystyle \frac{2b^2}{a}$.

Substitute $b^2 = 4$ and $a = \frac{6}{\sqrt{5}}$:

$\displaystyle \frac{2 \times 4}{6/\sqrt{5}} = \frac{8}{6/\sqrt{5}} = \frac{8}{1} \times \frac{\sqrt{5}}{6} = \frac{8\sqrt{5}}{6}$

Simplify the fraction:

$\displaystyle \frac{\cancel{8}^{4}\sqrt{5}}{\cancel{6}_{3}} = \frac{4\sqrt{5}}{3}$

---

Summary of Results:

Foci: $\left(0, \pm \frac{2\sqrt{70}}{5}\right)$ (or $\left(0, \pm \sqrt{\frac{56}{5}}\right)$)

Vertices: $\left(0, \pm \frac{6\sqrt{5}}{5}\right)$ (or $\left(0, \pm \sqrt{\frac{36}{5}}\right)$)

Eccentricity: $\frac{\sqrt{14}}{3}$

Length of Latus Rectum: $\frac{4\sqrt{5}}{3}$

Given:

The equation of the hyperbola is $49y^2 – 16x^2 = 784$.

---

To Find:

The coordinates of the foci, the vertices, the eccentricity, and the length of the latus rectum of the given hyperbola.

---

Solution:

The given equation of the hyperbola is:

$49y^2 – 16x^2 = 784$

To convert this to the standard form, divide the entire equation by 784:

$\displaystyle \frac{49y^2}{784} - \frac{16x^2}{784} = \frac{784}{784}$

Simplify the fractions:

$\displaystyle \frac{\cancel{49}y^2}{\cancel{784}_{16}} - \frac{\cancel{16}x^2}{\cancel{784}_{49}} = 1$

This equation is in the standard form $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$, where the centre is at the origin $(0, 0)$ and the transverse axis is along the y-axis (since the $y^2$ term is positive).

Comparing equation (i) with the standard form, we have:

$a^2 = 16 \implies a = \sqrt{16} = 4$ (since $a > 0$)

$b^2 = 49 \implies b = \sqrt{49} = 7$ (since $b > 0$)

Now we find the value of $c$ using the relation $c^2 = a^2 + b^2$ for a hyperbola:

$c^2 = 16 + 49$

$c^2 = 65$

$c = \sqrt{65}$ (since $c > 0$)

Using the values of $a, b, c$ and the fact that the transverse axis is along the y-axis:

1. Coordinates of the foci: The foci are at $(0, \pm c)$.

Substituting $c = \sqrt{65}$, the coordinates of the foci are $(0, \pm \sqrt{65})$, i.e., $(0, \sqrt{65})$ and $(0, -\sqrt{65})$.

2. Vertices: The vertices are at $(0, \pm a)$.

Substituting $a = 4$, the coordinates of the vertices are $(0, \pm 4)$, i.e., $(0, 4)$ and $(0, -4)$.

3. Eccentricity: The eccentricity $e$ is given by $e = \frac{c}{a}$.

Substitute $c = \sqrt{65}$ and $a = 4$, the eccentricity is $\displaystyle e = \frac{\sqrt{65}}{4}$.

4. Length of the latus rectum: The length of the latus rectum is $\displaystyle \frac{2b^2}{a}$.

Substitute $b^2 = 49$ and $a = 4$, the length of the latus rectum is $\displaystyle \frac{2 \times 49}{4}$.

Simplify the fraction:

$\displaystyle \frac{98}{4} = \frac{\cancel{98}^{49}}{\cancel{4}_{2}} = \frac{49}{2}$

---

Summary of Results:

Foci: $(0, \pm \sqrt{65})$

Vertices: $(0, \pm 4)$

Eccentricity: $\frac{\sqrt{65}}{4}$

Length of Latus Rectum: $\frac{49}{2}$

Given:

The vertices of the hyperbola are $(\pm 2, 0)$.

The foci of the hyperbola are $(\pm 3, 0)$.

---

To Find:

The equation of the hyperbola.

---

Solution:

The given vertices are $(\pm 2, 0)$ and the foci are $(\pm 3, 0)$.

Since the vertices and foci lie on the x-axis and are symmetric with respect to the origin, the centre of the hyperbola is at the origin $(0, 0)$, and the transverse axis is along the x-axis.

For a hyperbola with the transverse axis along the x-axis and centre at the origin, the standard form of the equation is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.

The coordinates of the vertices are $(\pm a, 0)$. Comparing with the given vertices $(\pm 2, 0)$, we have:

So, $a^2 = 2^2 = 4$.

The coordinates of the foci are $(\pm c, 0)$. Comparing with the given foci $(\pm 3, 0)$, we have:

So, $c^2 = 3^2 = 9$.

For a hyperbola, the relationship between $a$, $b$, and $c$ is $c^2 = a^2 + b^2$. We need to find $b^2$.

Rearrange the formula to solve for $b^2$:

$b^2 = c^2 - a^2$

Substitute the values of $c^2 = 9$ and $a^2 = 4$:

$b^2 = 9 - 4$

Now, substitute the values of $a^2 = 4$ and $b^2 = 5$ into the standard equation of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$:

This is the required equation of the hyperbola.

---

Final Equation:

The equation of the hyperbola is $\frac{x^2}{4} - \frac{y^2}{5} = 1$.

Given:

The vertices of the hyperbola are $(0, \pm 5)$.

The foci of the hyperbola are $(0, \pm 8)$.

---

To Find:

The equation of the hyperbola.

---

Solution:

The given vertices are $(0, \pm 5)$ and the foci are $(0, \pm 8)$.

Since the vertices and foci lie on the y-axis and are symmetric with respect to the origin, the centre of the hyperbola is at the origin $(0, 0)$, and the transverse axis is along the y-axis.

For a hyperbola with the transverse axis along the y-axis and centre at the origin, the standard form of the equation is $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$.

The coordinates of the vertices are $(0, \pm a)$. Comparing with the given vertices $(0, \pm 5)$, we have:

So, $a^2 = 5^2 = 25$.

The coordinates of the foci are $(0, \pm c)$. Comparing with the given foci $(0, \pm 8)$, we have:

So, $c^2 = 8^2 = 64$.

For a hyperbola, the relationship between $a$, $b$, and $c$ is $c^2 = a^2 + b^2$. We need to find $b^2$.

Rearrange the formula to solve for $b^2$:

$b^2 = c^2 - a^2$

Substitute the values of $c^2 = 64$ and $a^2 = 25$:

$b^2 = 64 - 25$

Now, substitute the values of $a^2 = 25$ and $b^2 = 39$ into the standard equation of the hyperbola $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$:

This is the required equation of the hyperbola.

---

Final Equation:

The equation of the hyperbola is $\frac{y^2}{25} - \frac{x^2}{39} = 1$.

Given:

The vertices of the hyperbola are $(0, \pm 3)$.

The foci of the hyperbola are $(0, \pm 5)$.

---

To Find:

The equation of the hyperbola.

---

Solution:

The given vertices are $(0, \pm 3)$ and the foci are $(0, \pm 5)$.

Since the vertices and foci lie on the y-axis and are symmetric with respect to the origin, the centre of the hyperbola is at the origin $(0, 0)$, and the transverse axis is along the y-axis.

For a hyperbola with the transverse axis along the y-axis and centre at the origin, the standard form of the equation is $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$.

The coordinates of the vertices are $(0, \pm a)$. Comparing with the given vertices $(0, \pm 3)$, we have:

So, $a^2 = 3^2 = 9$.

The coordinates of the foci are $(0, \pm c)$. Comparing with the given foci $(0, \pm 5)$, we have:

So, $c^2 = 5^2 = 25$.

For a hyperbola, the relationship between $a$, $b$, and $c$ is $c^2 = a^2 + b^2$. We need to find $b^2$.

Rearrange the formula to solve for $b^2$:

$b^2 = c^2 - a^2$

Substitute the values of $c^2 = 25$ and $a^2 = 9$:

$b^2 = 25 - 9$

Now, substitute the values of $a^2 = 9$ and $b^2 = 16$ into the standard equation of the hyperbola $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$:

This is the required equation of the hyperbola.

---

Final Equation:

The equation of the hyperbola is $\frac{y^2}{9} - \frac{x^2}{16} = 1$.

Given:

The foci of the hyperbola are $(\pm 5, 0)$.

The length of the transverse axis is 8.

---

To Find:

The equation of the hyperbola.

---

Solution:

The given foci are $(\pm 5, 0)$. Since the foci lie on the x-axis and are symmetric with respect to the origin, the centre of the hyperbola is at the origin $(0, 0)$, and the transverse axis is along the x-axis.

For a hyperbola with the transverse axis along the x-axis and centre at the origin, the standard form of the equation is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.

The coordinates of the foci are $(\pm c, 0)$. Comparing with the given foci $(\pm 5, 0)$, we have:

So, $c^2 = 5^2 = 25$.

The length of the transverse axis is given as 8. The length of the transverse axis is equal to $2a$.

$\displaystyle 2a = 8$

Divide by 2:

So, $a^2 = 4^2 = 16$.

For a hyperbola, the relationship between $a$, $b$, and $c$ is $c^2 = a^2 + b^2$. We need to find $b^2$.

Rearrange the formula to solve for $b^2$:

$b^2 = c^2 - a^2$

Substitute the values of $c^2 = 25$ and $a^2 = 16$:

$b^2 = 25 - 16$

Now, substitute the values of $a^2 = 16$ and $b^2 = 9$ into the standard equation of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$:

This is the required equation of the hyperbola.

---

Final Equation:

The equation of the hyperbola is $\frac{x^2}{16} - \frac{y^2}{9} = 1$.

Given:

The foci of the hyperbola are $(0, \pm 13)$.

The length of the conjugate axis is 24.

---

To Find:

The equation of the hyperbola.

---

Solution:

The given foci are $(0, \pm 13)$. Since the foci lie on the y-axis and are symmetric with respect to the origin, the centre of the hyperbola is at the origin $(0, 0)$, and the transverse axis is along the y-axis.

For a hyperbola with the transverse axis along the y-axis and centre at the origin, the standard form of the equation is $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$.

The coordinates of the foci are $(0, \pm c)$. Comparing with the given foci $(0, \pm 13)$, we have:

So, $c^2 = 13^2 = 169$.

The length of the conjugate axis is given as 24. The length of the conjugate axis is equal to $2b$.

$\displaystyle 2b = 24$

Divide by 2:

So, $b^2 = 12^2 = 144$.

For a hyperbola, the relationship between $a$, $b$, and $c$ is $c^2 = a^2 + b^2$. We need to find $a^2$.

Rearrange the formula to solve for $a^2$:

$a^2 = c^2 - b^2$

Substitute the values of $c^2 = 169$ and $b^2 = 144$:

$a^2 = 169 - 144$

Now, substitute the values of $a^2 = 25$ and $b^2 = 144$ into the standard equation of the hyperbola $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$:

This is the required equation of the hyperbola.

---

Final Equation:

The equation of the hyperbola is $\frac{y^2}{25} - \frac{x^2}{144} = 1$.

Given:

The foci of the hyperbola are $(\pm 3\sqrt{5} , 0)$.

The length of the latus rectum is 8.

---

To Find:

The equation of the hyperbola.

---

Solution:

The given foci are $(\pm 3\sqrt{5} , 0)$. Since the foci lie on the x-axis and are symmetric with respect to the origin, the centre of the hyperbola is at the origin $(0, 0)$, and the transverse axis is along the x-axis.

For a hyperbola with the transverse axis along the x-axis and centre at the origin, the standard form of the equation is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.

The coordinates of the foci are $(\pm c, 0)$. Comparing with the given foci $(\pm 3\sqrt{5}, 0)$, we have:

So, $c^2 = (3\sqrt{5})^2 = 3^2 \times (\sqrt{5})^2 = 9 \times 5 = 45$.

The length of the latus rectum is given as 8. The formula for the length of the latus rectum of a hyperbola is $\frac{2b^2}{a}$.

$\displaystyle \frac{2b^2}{a} = 8$

From this equation, we can express $b^2$ in terms of $a$:

$2b^2 = 8a$

For a hyperbola, the relationship between $a$, $b$, and $c$ is $c^2 = a^2 + b^2$.

Substitute the value of $c^2 = 45$ into this equation:

Now, substitute the expression for $b^2$ from equation (1) into equation (2):

$45 = a^2 + 4a$

Rearrange this into a quadratic equation in $a$:

$a^2 + 4a - 45 = 0$

We can solve this quadratic equation for $a$ by factoring or using the quadratic formula. We need two numbers that multiply to -45 and add up to 4. These numbers are 9 and -5.

$(a + 9)(a - 5) = 0$

This gives two possible values for $a$:

$a + 9 = 0 \implies a = -9$

$a - 5 = 0 \implies a = 5$

Since $a$ represents a distance (half the length of the transverse axis), it must be positive. Therefore, $a = 5$.

Now, substitute the value of $a = 5$ back into equation (1) to find $b^2$:

$b^2 = 4a$

$b^2 = 4(5)$

We have $a = 5$ (so $a^2 = 25$) and $b^2 = 20$.

For a hyperbola with the transverse axis along the x-axis and centre at the origin, the standard form of the equation is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.

Substitute the values of $a^2 = 25$ and $b^2 = 20$ into the standard equation:

This is the required equation of the hyperbola.

---

Final Equation:

The equation of the hyperbola is $\frac{x^2}{25} - \frac{y^2}{20} = 1$.

Given:

The foci of the hyperbola are $(\pm 4, 0)$.

The length of the latus rectum is 12.

---

To Find:

The equation of the hyperbola.

---

Solution:

The given foci are $(\pm 4, 0)$. Since the foci lie on the x-axis and are symmetric with respect to the origin, the centre of the hyperbola is at the origin $(0, 0)$, and the transverse axis is along the x-axis.

For a hyperbola with the transverse axis along the x-axis and centre at the origin, the standard form of the equation is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.

The coordinates of the foci are $(\pm c, 0)$. Comparing with the given foci $(\pm 4, 0)$, we have:

So, $c^2 = 4^2 = 16$.

The length of the latus rectum is given as 12. The formula for the length of the latus rectum of a hyperbola is $\frac{2b^2}{a}$.

$\displaystyle \frac{2b^2}{a} = 12$

From this equation, we can express $b^2$ in terms of $a$:

$2b^2 = 12a$

For a hyperbola, the relationship between $a$, $b$, and $c$ is $c^2 = a^2 + b^2$.

Substitute the value of $c^2 = 16$ into this equation:

Now, substitute the expression for $b^2$ from equation (1) into equation (2):

$16 = a^2 + 6a$

Rearrange this into a quadratic equation in $a$:

$a^2 + 6a - 16 = 0$

We can solve this quadratic equation for $a$ by factoring or using the quadratic formula. We need two numbers that multiply to -16 and add up to 6. These numbers are 8 and -2.

$(a + 8)(a - 2) = 0$

This gives two possible values for $a$:

$a + 8 = 0 \implies a = -8$

$a - 2 = 0 \implies a = 2$

Since $a$ represents a distance (half the length of the transverse axis), it must be positive. Therefore, $a = 2$.

Now, substitute the value of $a = 2$ back into equation (1) to find $b^2$:

$b^2 = 6a$

$b^2 = 6(2)$

We have $a = 2$ (so $a^2 = 4$) and $b^2 = 12$.

For a hyperbola with the transverse axis along the x-axis and centre at the origin, the standard form of the equation is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.

Substitute the values of $a^2 = 4$ and $b^2 = 12$ into the standard equation:

This is the required equation of the hyperbola.

---

Final Equation:

The equation of the hyperbola is $\frac{x^2}{4} - \frac{y^2}{12} = 1$.

Given:

The vertices of the hyperbola are $(\pm 7, 0)$.

The eccentricity of the hyperbola is $e = \frac{4}{3}$.

---

To Find:

The equation of the hyperbola.

---

Solution:

The given vertices are $(\pm 7, 0)$. Since the vertices lie on the x-axis and are symmetric with respect to the origin, the centre of the hyperbola is at the origin $(0, 0)$, and the transverse axis is along the x-axis.

For a hyperbola with the transverse axis along the x-axis and centre at the origin, the standard form of the equation is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.

The coordinates of the vertices are $(\pm a, 0)$. Comparing with the given vertices $(\pm 7, 0)$, we have:

So, $a^2 = 7^2 = 49$.

The eccentricity is given by $e = \frac{4}{3}$. The formula for the eccentricity of a hyperbola is $e = \frac{c}{a}$.

$\displaystyle \frac{c}{a} = \frac{4}{3}$

Substitute the value of $a = 7$:

$\displaystyle \frac{c}{7} = \frac{4}{3}$

Multiply both sides by 7 to find $c$:

$\displaystyle c = \frac{4}{3} \times 7 = \frac{28}{3}$

So, $c^2 = \left(\frac{28}{3}\right)^2 = \frac{784}{9}$.

For a hyperbola, the relationship between $a$, $b$, and $c$ is $c^2 = a^2 + b^2$. We need to find $b^2$.

Rearrange the formula to solve for $b^2$:

$b^2 = c^2 - a^2$

Substitute the values of $c^2 = \frac{784}{9}$ and $a^2 = 49$:

$\displaystyle b^2 = \frac{784}{9} - 49$

Find a common denominator:

$\displaystyle b^2 = \frac{784 - 49 \times 9}{9} = \frac{784 - 441}{9} = \frac{343}{9}$

Now, substitute the values of $a^2 = 49$ and $b^2 = \frac{343}{9}$ into the standard equation of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$:

$\displaystyle \frac{x^2}{49} - \frac{y^2}{343/9} = 1$

Rewrite the second term:

This is the required equation of the hyperbola.

---

Final Equation:

The equation of the hyperbola is $\frac{x^2}{49} - \frac{9y^2}{343} = 1$.

Given:

The foci of the hyperbola are $(0, \pm \sqrt{10})$.

The hyperbola passes through the point $(2, 3)$.

---

To Find:

The equation of the hyperbola.

---

Solution:

The given foci are $(0, \pm \sqrt{10})$. Since the foci lie on the y-axis and are symmetric with respect to the origin, the centre of the hyperbola is at the origin $(0, 0)$, and the transverse axis is along the y-axis.

For a hyperbola with the transverse axis along the y-axis and centre at the origin, the standard form of the equation is $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$.

The coordinates of the foci are $(0, \pm c)$. Comparing with the given foci $(0, \pm \sqrt{10})$, we have:

So, $c^2 = (\sqrt{10})^2 = 10$.

For a hyperbola, the relationship between $a$, $b$, and $c$ is $c^2 = a^2 + b^2$.

Substitute the value of $c^2 = 10$:

We can express $b^2$ in terms of $a^2$ from equation (1):

Since the hyperbola passes through the point $(2, 3)$, substitute $x = 2$ and $y = 3$ into the standard equation $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$:

Substitute the expression for $b^2$ from equation (2) into equation (3):

$\displaystyle \frac{9}{a^2} - \frac{4}{10 - a^2} = 1$

To eliminate the denominators, multiply the entire equation by $a^2(10 - a^2)$:

$\displaystyle a^2(10 - a^2)\left(\frac{9}{a^2}\right) - a^2(10 - a^2)\left(\frac{4}{10 - a^2}\right) = a^2(10 - a^2)(1)$

$\displaystyle \cancel{a^2}(10 - a^2)\left(\frac{9}{\cancel{a^2}}\right) - a^2\cancel{(10 - a^2)}\left(\frac{4}{\cancel{10 - a^2}}\right) = a^2(10 - a^2)$

$9(10 - a^2) - 4a^2 = a^2(10 - a^2)$

$90 - 9a^2 - 4a^2 = 10a^2 - (a^2)^2$

$90 - 13a^2 = 10a^2 - a^4$

Move all terms to one side to form a quadratic equation in $a^2$:

$a^4 - 10a^2 - 13a^2 + 90 = 0$

$a^4 - 23a^2 + 90 = 0$

Let $A = a^2$. The equation becomes a quadratic equation in $A$:

$A^2 - 23A + 90 = 0$

We can solve this quadratic equation for $A$ by factoring or using the quadratic formula. We need two numbers that multiply to 90 and add up to -23. These numbers are -18 and -5.

$(A - 18)(A - 5) = 0$

This gives two possible values for $A$ (and thus for $a^2$):

$A - 18 = 0 \implies A = 18 \implies a^2 = 18$

$A - 5 = 0 \implies A = 5 \implies a^2 = 5$

Now we find the corresponding values of $b^2$ using $b^2 = 10 - a^2$ (from equation 2).

If $a^2 = 18$, then $b^2 = 10 - 18 = -8$. This is not possible since $b^2$ must be positive.

If $a^2 = 5$, then $b^2 = 10 - 5 = 5$. This is possible since $b^2$ is positive.

So, we have $a^2 = 5$ and $b^2 = 5$.

The standard equation of the hyperbola is $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$. Substitute the values of $a^2 = 5$ and $b^2 = 5$:

Multiply the entire equation by 5 to clear the denominator:

This is the required equation of the hyperbola.

---

Final Equation:

The equation of the hyperbola is $\frac{y^2}{5} - \frac{x^2}{5} = 1$, or $y^2 - x^2 = 5$.

Given:

A parabolic mirror with its vertex at the origin.

The distance from the vertex to the focus is 5 cm.

The depth of the mirror is 45 cm.

---

To Find:

The distance AB, which represents the diameter of the mirror at its edge.

---

Solution:

As shown in the figure, the vertex of the parabolic mirror is at the origin (0, 0), and the parabola opens to the right. The axis of symmetry is along the x-axis.

The standard equation for a parabola with its vertex at the origin and opening to the right is:

The focus of this parabola is at the point $(a, 0)$. We are given that the distance of the focus from the vertex is 5 cm. Therefore, the value of $a$ is:

$a = 5$ cm

Substituting this value of $a$ into the standard equation (i), we get the specific equation for this parabolic mirror:

$y^2 = 4(5)x$

We are given that the mirror is 45 cm deep. This depth corresponds to the x-coordinate of the points A and B on the edge of the mirror.

So, for points A and B, we have $x = 45$ cm.

To find the corresponding y-coordinates, we substitute $x = 45$ into the equation of the parabola (ii):

$y^2 = 20(45)$

$y^2 = 900$

Taking the square root of both sides:

$y = \pm \sqrt{900}$

$y = \pm 30$

This means the coordinates of the endpoints are A(45, 30) and B(45, -30).

The distance AB is the vertical distance between these two points.

Distance AB = (y-coordinate of A) - (y-coordinate of B)

Distance AB = $30 - (-30)$

Distance AB = $30 + 30 = 60$ cm.

---

Final Answer:

The distance AB is 60 cm.

Given:

A beam supported at its ends, with the deflected shape being a parabola.

The distance between the supports is 12 metres.

The maximum deflection at the centre is 3 cm.

---

To Find:

The horizontal distance from the centre where the deflection is 1 cm.

---

Solution:

Let's set up a coordinate system to model the parabolic beam. We can place the vertex of the parabola at the origin (0, 0). Since the beam deflects downwards, we will consider a parabola that opens upwards.

The standard equation for an upward-opening parabola with its vertex at the origin is:

The supports are 12 metres apart, and the load is concentrated at the centre. With the vertex (centre) at $x=0$, the supports are located at $x = -6$ m and $x = 6$ m.

The maximum deflection at the centre is 3 cm, which is 0.03 metres. This means the supports are 0.03 m higher than the lowest point (the vertex). Therefore, the coordinates of the supports are $(-6, 0.03)$ and $(6, 0.03)$.

To find the equation of this specific parabola, we can substitute the coordinates of one of the supports, say $(6, 0.03)$, into the standard equation (i):

$(6)^2 = 4a(0.03)$

$36 = 0.12a$

Now, we solve for $a$:

$a = \frac{36}{0.12} = \frac{3600}{12} = 300$

Substituting the value of $a$ back into the standard equation (i), we get the equation of the parabolic beam:

$x^2 = 4(300)y$

We need to find the horizontal distance from the centre, $x$, where the deflection is 1 cm.

Deflection is measured from the horizontal line connecting the supports (the undeflected position). The supports are at a height of $y = 3$ cm. A deflection of 1 cm means the point on the beam is 1 cm below the level of the supports.

So, the height ($y$-coordinate) of this point is:

$y = (\text{Height of supports}) - (\text{Deflection})$

$y = 3 \text{ cm} - 1 \text{ cm} = 2 \text{ cm} = 0.02 \text{ m}$

Now, we substitute $y = 0.02$ into the equation of our parabola (ii) to find the corresponding value of $x$:

$x^2 = 1200(0.02)$

$x^2 = 24$

Taking the square root of both sides:

$x = \pm \sqrt{24} = \pm \sqrt{4 \times 6} = \pm 2\sqrt{6}$

The distance from the centre is the absolute value of $x$, which is $2\sqrt{6}$ metres.

---

Final Answer:

The deflection is 1 cm at a distance of $2\sqrt{6}$ metres from the centre of the beam.

Given:

A rod AB of length 15 cm moves with its end A on the x-axis and end B on the y-axis.

A point P(x, y) is on the rod such that the distance from end A is AP = 6 cm.

---

To Prove:

The locus of the point P(x, y) is an ellipse.

---

Proof:

Let the rod AB make an angle $\theta$ with the positive x-axis at any given instant. The point P has coordinates $(x, y)$.

The total length of the rod is AB = 15 cm.

We are given that AP = 6 cm.

Therefore, the length of the remaining part of the rod is PB = AB - AP = 15 - 6 = 9 cm.

From the point P(x, y), draw a line PM perpendicular to the x-axis and a line PN perpendicular to the y-axis. This means PM = y and PN = x.

Now, consider the right-angled triangle formed with hypotenuse PB. The line PN is parallel to the x-axis, so the angle BPN is equal to $\theta$. In this triangle:

$\cos\theta = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{PN}{PB} = \frac{x}{9}$

Next, consider the right-angled triangle PMA. The angle at A is $\theta$. In this triangle:

$\sin\theta = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{PM}{AP} = \frac{y}{6}$

From equations (i) and (ii), we can express $\cos\theta$ and $\sin\theta$ in terms of x and y:

$\cos\theta = \frac{x}{9}$

$\sin\theta = \frac{y}{6}$

We know the fundamental trigonometric identity:

Substitute the expressions for $\cos\theta$ and $\sin\theta$ into this identity to eliminate $\theta$ and find the equation for the locus of P:

$\left(\frac{x}{9}\right)^2 + \left(\frac{y}{6}\right)^2 = 1$

This is the standard equation of an ellipse with its centre at the origin, a semi-major axis $a=9$ along the x-axis, and a semi-minor axis $b=6$ along the y-axis.

Hence, the locus of the point P(x, y) is an ellipse.

---

Alternate Solution (Using Section Formula):

Let the coordinates of the end A on the x-axis be $(a, 0)$ and the end B on the y-axis be $(0, b)$.

Since the length of the rod AB is 15 cm, by the distance formula:

$\sqrt{(a-0)^2 + (0-b)^2} = 15$

The point P(x, y) is on the rod such that AP = 6 cm and PB = 15 - 6 = 9 cm. So, P divides the line segment BA in the ratio BP:PA = 9:6, which simplifies to 3:2.

Using the section formula with endpoints B(0,b) and A(a,0):

$x = \frac{2 \cdot 0 + 3 \cdot a}{3+2} = \frac{3a}{5} \implies a = \frac{5x}{3}$

$y = \frac{2 \cdot b + 3 \cdot 0}{3+2} = \frac{2b}{5} \implies b = \frac{5y}{2}$

Now, substitute these expressions for $a$ and $b$ into equation (A):

$\left(\frac{5x}{3}\right)^2 + \left(\frac{5y}{2}\right)^2 = 225$

$\frac{25x^2}{9} + \frac{25y^2}{4} = 225$

Divide the entire equation by 225 (since $225 = 9 \times 25$):

$\frac{25x^2}{9 \cdot 225} + \frac{25y^2}{4 \cdot 225} = 1$

$\frac{\cancel{25}x^2}{9 \cdot \cancel{25} \cdot 9} + \frac{\cancel{25}y^2}{4 \cdot \cancel{25} \cdot 9} = 1$

$\frac{x^2}{81} + \frac{y^2}{36} = 1$

This is the equation of an ellipse. Hence, the locus of P is an ellipse.

Given:

Diameter of the parabolic reflector = 20 cm.

Depth of the parabolic reflector = 5 cm.

---

To Find:

The focus of the reflector.

---

Solution:

Let's orient the parabolic reflector on a coordinate plane. We can place the vertex of the parabola at the origin (0, 0) and have the axis of symmetry along the positive x-axis. This means the parabola opens to the right.

The standard equation for such a parabola is:

The focus of this parabola is at the point F(a, 0).

The depth of the reflector is 5 cm, which corresponds to the x-coordinate of the rim. The diameter is 20 cm, which means the radius is 10 cm. This radius corresponds to the y-coordinate of the rim.

So, a point on the rim of the reflector has coordinates (5, 10).

Since this point lies on the parabola, it must satisfy the equation of the parabola. We substitute $x = 5$ and $y = 10$ into equation (i):

$(10)^2 = 4a(5)$

$100 = 20a$

Solving for $a$:

$a = \frac{100}{20} = 5$

The focus is at the point $(a, 0)$. Substituting the value of $a$, the focus is at $(5, 0)$.

This means the focus is located on the axis of symmetry, 5 cm away from the vertex, inside the reflector.

---

Final Answer:

The focus is at a distance of 5 cm from the vertex.

Given:

A parabolic arch with a vertical axis.

Height of the arch = 10 m.

Width of the arch at the base = 5 m.

---

To Find:

The width of the arch at a distance of 2 m from the vertex.

---

Solution:

Let's set up a coordinate system. Since the arch has a vertical axis, we can place its vertex at the origin (0, 0). As it's an arch, the parabola opens downwards.

The standard equation for a downward-opening parabola with its vertex at the origin is:

The arch is 10 m high. This means the base is 10 m below the vertex, so the y-coordinate of the base is $y = -10$.

The arch is 5 m wide at the base. Due to symmetry about the y-axis, the x-coordinates of the base endpoints are $x = \pm \frac{5}{2} = \pm 2.5$.

So, the point $(2.5, -10)$ lies on the parabola.

Substitute these coordinates into the standard equation (i) to find the value of $a$:

$(2.5)^2 = -4a(-10)$

$6.25 = 40a$

$a = \frac{6.25}{40} = \frac{625}{4000} = \frac{1}{6.4} = \frac{5}{32}$

Now, substitute the value of $4a = 4(\frac{5}{32}) = \frac{5}{8}$ back into the equation (i):

We need to find the width of the arch 2 m from the vertex. This means we need to find the width at a height 2 m below the vertex, which corresponds to the y-coordinate $y = -2$.

Substitute $y = -2$ into equation (ii):

$x^2 = -\frac{5}{8}(-2)$

$x^2 = \frac{10}{8} = \frac{5}{4}$

Solving for $x$:

$x = \pm \sqrt{\frac{5}{4}} = \pm \frac{\sqrt{5}}{2}$

The two points on the arch are $(\frac{\sqrt{5}}{2}, -2)$ and $(-\frac{\sqrt{5}}{2}, -2)$.

The width of the arch at this height is the distance between these two points:

Width = $\frac{\sqrt{5}}{2} - \left(-\frac{\sqrt{5}}{2}\right) = \frac{\sqrt{5}}{2} + \frac{\sqrt{5}}{2} = \sqrt{5}$ metres.

---

Final Answer:

The width of the arch 2 m from the vertex is $\sqrt{5}$ metres.

Given:

A suspension bridge cable in the shape of a parabola.

Length of the horizontal roadway = 100 m.

Length of the longest supporting wire = 30 m.

Length of the shortest supporting wire = 6 m.

---

To Find:

The length of a supporting wire 18 m from the middle of the roadway.

---

Solution:

Let's set up a coordinate system. The shortest wire is at the middle of the bridge, which corresponds to the vertex of the parabola. Let the vertex of the parabola be at the point $(0, 6)$, placing the horizontal roadway along the x-axis ($y=0$). The parabola opens upwards.

The standard equation for an upward-opening parabola with vertex at $(h, k)$ is:

Substituting the vertex $(h, k) = (0, 6)$, the equation becomes:

The roadway is 100 m long, so it extends from $x = -50$ m to $x = 50$ m. The longest supporting wires are at the ends of the roadway. The length of these wires is 30 m. This means the cable passes through the points $(\pm 50, 30)$.

We can use one of these points, say $(50, 30)$, to find the value of $a$. Substitute $x=50$ and $y=30$ into equation (i):

$(50)^2 = 4a(30 - 6)$

$2500 = 4a(24)$

$2500 = 96a$

$a = \frac{2500}{96} = \frac{625}{24}$

Now, substitute the value of $4a = 4(\frac{625}{24}) = \frac{625}{6}$ back into equation (i) to get the specific equation for the cable:

We need to find the length of the supporting wire 18 m from the middle. This corresponds to a horizontal position of $x = 18$ m. The length of the wire is the y-coordinate of the cable at this position.

Substitute $x=18$ into equation (ii) and solve for $y$:

$(18)^2 = \frac{625}{6}(y-6)$

$324 = \frac{625}{6}(y-6)$

To isolate $(y-6)$, multiply both sides by $\frac{6}{625}$:

$y-6 = 324 \times \frac{6}{625}$

$y-6 = \frac{1944}{625}$

$y = 6 + \frac{1944}{625}$

$y = \frac{6 \times 625 + 1944}{625} = \frac{3750 + 1944}{625} = \frac{5694}{625}$

The length of the wire is $y = \frac{5694}{625} = 9.1104$ metres.

---

Final Answer:

The length of the supporting wire 18 m from the middle is 9.11 m (approx).

Given:

A semi-elliptical arch.

Width of the arch at the base = 8 m.

Height of the arch at the centre = 2 m.

---

To Find:

The height of the arch at a point 1.5 m from one end.

---

Solution:

Let's model the semi-elliptical arch on a coordinate plane. We can place the centre of the ellipse at the origin (0, 0) and the wide base along the x-axis.

The width of the arch is 8 m. This corresponds to the length of the major axis of the full ellipse. The semi-major axis length is half of this width:

$a = \frac{8}{2} = 4$ m

The height of the arch at the centre is 2 m. This corresponds to the semi-minor axis length:

$b = 2$ m

Since the base is along the x-axis, the standard equation of the ellipse is:

Substituting the values of $a=4$ and $b=2$:

We need to find the height of the arch at a point 1.5 m from one end. The ends of the arch are at $x = -4$ and $x = 4$.

A point 1.5 m from the end at $x=4$ will have an x-coordinate of:

$x = 4 - 1.5 = 2.5$ m

Now, we need to find the height, $y$, at this x-coordinate. Substitute $x=2.5$ into the ellipse equation (i):

$\frac{(2.5)^2}{16} + \frac{y^2}{4} = 1$

$\frac{6.25}{16} + \frac{y^2}{4} = 1$

Isolate the $y^2$ term:

$\frac{y^2}{4} = 1 - \frac{6.25}{16}$

$\frac{y^2}{4} = \frac{16 - 6.25}{16} = \frac{9.75}{16}$

Solve for $y^2$:

$y^2 = 4 \times \frac{9.75}{16} = \frac{9.75}{4}$

Solve for $y$ by taking the positive square root (since height is positive):

$y = \sqrt{\frac{9.75}{4}} = \frac{\sqrt{9.75}}{2}$

We can also write $9.75$ as $\frac{39}{4}$, so $y^2 = \frac{39/4}{4} = \frac{39}{16}$.

$y = \sqrt{\frac{39}{16}} = \frac{\sqrt{39}}{4}$

Calculating the approximate value: $\sqrt{39} \approx 6.245$.

$y \approx \frac{6.245}{4} \approx 1.56$ m.

---

Final Answer:

The height of the arch at a point 1.5 m from one end is $\frac{\sqrt{39}}{4}$ metres (approximately 1.56 m).

Given:

A rod of length 12 cm moves with its ends on the x and y axes.

A point P(x, y) on the rod is 3 cm from the end in contact with the x-axis.

---

To Determine:

The equation of the locus of point P.

---

Solution:

Let the rod be represented by the line segment AB, where the end A lies on the x-axis and the end B lies on the y-axis. The length of the rod is AB = 12 cm.

Let the coordinates of A be $(h, 0)$ and the coordinates of B be $(0, k)$.

Since the length of the rod is 12 cm, the distance between A and B is always 12. Using the distance formula:

$\sqrt{(h - 0)^2 + (0 - k)^2} = 12$

Squaring both sides, we get the relationship between h and k:

The point P(x, y) is on the rod and is 3 cm from the end A (which is in contact with the x-axis). Therefore, AP = 3 cm.

The length of the other part of the rod is PB = AB - AP = 12 - 3 = 9 cm.

So, the point P divides the line segment AB in the ratio AP : PB = 3 : 9, which simplifies to 1 : 3.

We can find the coordinates of P(x, y) using the section formula, which states that if a point divides the line segment joining $(x_1, y_1)$ and $(x_2, y_2)$ in the ratio $m:n$, its coordinates are $(\frac{mx_2+nx_1}{m+n}, \frac{my_2+ny_1}{m+n})$.

Here, $(x_1, y_1) = A(h, 0)$, $(x_2, y_2) = B(0, k)$, and the ratio $m:n = 1:3$.

$x = \frac{1(0) + 3(h)}{1 + 3} = \frac{3h}{4}$

$y = \frac{1(k) + 3(0)}{1 + 3} = \frac{k}{4}$

From these relations, we can express $h$ and $k$ in terms of $x$ and $y$ to eliminate them:

$h = \frac{4x}{3}$

$k = 4y$

Now, substitute these expressions for $h$ and $k$ into equation (i):

$\left(\frac{4x}{3}\right)^2 + (4y)^2 = 144$

$\frac{16x^2}{9} + 16y^2 = 144$

To obtain the standard form of the equation, divide the entire equation by 144:

$\frac{16x^2}{9 \times 144} + \frac{16y^2}{144} = \frac{144}{144}$

$\frac{x^2}{9 \times 9} + \frac{y^2}{9} = 1$

This is the equation of an ellipse, which is the locus of the point P.

---

Final Answer:

The equation of the locus of point P is $\frac{x^2}{81} + \frac{y^2}{9} = 1$.

Given:

The equation of the parabola is $x^2 = 12y$.

---

To Find:

The area of the triangle formed by joining the vertex to the ends of the latus rectum.

---

Solution:

The given equation of the parabola is $x^2 = 12y$.

This is in the standard form $x^2 = 4ay$, which represents a parabola opening upwards with its vertex at the origin.

1. Find the Vertex and Focus:

The vertex of the parabola is at the origin, $V(0, 0)$.

Comparing $x^2 = 12y$ with $x^2 = 4ay$, we get:

$4a = 12 \implies a = 3$

The focus of the parabola is at $F(0, a)$, which is $F(0, 3)$.

2. Find the Endpoints of the Latus Rectum:

The latus rectum is a line segment that passes through the focus, is parallel to the directrix, and has its endpoints on the parabola. For this parabola, the latus rectum lies on the horizontal line $y = 3$.

To find the x-coordinates of the endpoints, substitute $y = 3$ into the parabola's equation:

$x^2 = 12(3) = 36$

$x = \pm \sqrt{36} = \pm 6$

So, the endpoints of the latus rectum are $L(-6, 3)$ and $R(6, 3)$.

3. Calculate the Area of the Triangle:

The triangle has vertices $V(0, 0)$, $L(-6, 3)$, and $R(6, 3)$.

We can use the formula: Area = $\frac{1}{2} \times \text{base} \times \text{height}$.

Let the base of the triangle be the latus rectum, the segment LR. The length of the base is the distance between $(-6, 3)$ and $(6, 3)$:

Base = $|6 - (-6)| = 12$ units.

(Note: The length of the latus rectum is also given by the formula $4a = 4(3) = 12$).

The height of the triangle is the perpendicular distance from the vertex $V(0, 0)$ to the base line $y=3$.

Height = $3$ units.

Now, calculate the area:

Area = $\frac{1}{2} \times 12 \times 3 = 18$ square units.

---

Final Answer:

The area of the triangle is 18 square units.

Given:

The sum of the distances of a man from two fixed flag posts is always constant and equal to 10 m.

The distance between the two flag posts is 8 m.

---

To Find:

The equation of the path traced by the man.

---

Solution:

The problem describes the locus of a point (the man) where the sum of its distances from two fixed points (the flag posts) is constant. By definition, this locus is an ellipse.

The two fixed flag posts are the foci of the ellipse.

Let's set up a coordinate system. We can place the two foci on the x-axis, symmetric with respect to the origin. The centre of the ellipse will be at (0, 0).

The distance between the foci is given as 8 m. This distance is equal to $2c$.

$2c = 8 \implies c = 4$ m

So, the foci are located at $F_1(-4, 0)$ and $F_2(4, 0)$.

The sum of the distances from any point on the ellipse to the two foci is constant and is equal to the length of the major axis, $2a$. We are given this sum is 10 m.

$2a = 10 \implies a = 5$ m

Since the foci are on the x-axis, the major axis is along the x-axis. The standard equation for such an ellipse is:

We have $a=5$ and $c=4$. For an ellipse, the relationship between $a, b,$ and $c$ is $c^2 = a^2 - b^2$. We can find $b^2$:

$b^2 = a^2 - c^2$

$b^2 = 5^2 - 4^2$

$b^2 = 25 - 16 = 9$

Now, we substitute the values of $a^2=25$ and $b^2=9$ into the standard equation:

This is the equation of the elliptical path traced by the man.

---

Final Answer:

The equation of the path is $\frac{x^2}{25} + \frac{y^2}{9} = 1$.

Given:

An equilateral triangle is inscribed in the parabola $y^2 = 4ax$.

One vertex of the triangle is at the vertex of the parabola.

---

To Find:

The length of the side of the equilateral triangle.

---

Solution:

The equation of the parabola is $y^2 = 4ax$. Its vertex is at the origin, $O(0, 0)$. Let this be one vertex of the equilateral triangle OAB.

Since the parabola is symmetric about the x-axis, and the triangle OAB has a vertex at O, the other two vertices A and B must be symmetric with respect to the x-axis to form an equilateral triangle.

Let the coordinates of vertex A be $(x, y)$. Then the coordinates of vertex B will be $(x, -y)$.

Since OAB is an equilateral triangle, all its angles are $60^\circ$. The x-axis bisects the angle $\angle AOB$. Therefore, the angle $\angle XOA$ is $30^\circ$.

In the right-angled triangle formed by dropping a perpendicular from A to the x-axis, we can relate the coordinates $(x,y)$ using trigonometry:

$\tan(30^\circ) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{y}{x}$

We know that $\tan(30^\circ) = \frac{1}{\sqrt{3}}$.

$\frac{y}{x} = \frac{1}{\sqrt{3}} \implies x = y\sqrt{3}$

Since the point A(x, y) lies on the parabola $y^2 = 4ax$, it must satisfy the equation. Substitute $x = y\sqrt{3}$ into the parabola's equation:

$y^2 = 4a(y\sqrt{3})$

Since A is a vertex different from the origin, $y \neq 0$. We can divide both sides by $y$:

$y = 4a\sqrt{3}$

Now we can find the corresponding x-coordinate:

$x = y\sqrt{3} = (4a\sqrt{3})\sqrt{3} = 4a(3) = 12a$

So, the coordinates of vertex A are $(12a, 4a\sqrt{3})$.

The length of the side of the triangle, let's say OA, can be found using the distance formula from the origin $O(0,0)$ to $A(12a, 4a\sqrt{3})$:

Length of side = $\sqrt{(12a-0)^2 + (4a\sqrt{3}-0)^2}$

Length of side = $\sqrt{(12a)^2 + (4a\sqrt{3})^2}$

Length of side = $\sqrt{144a^2 + 16a^2 \cdot 3}$

Length of side = $\sqrt{144a^2 + 48a^2}$

Length of side = $\sqrt{192a^2}$

Length of side = $\sqrt{64 \times 3 \times a^2}$

Length of side = $8\sqrt{3}a$

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Final Answer:

The length of the side of the equilateral triangle is $8\sqrt{3}a$.