Chapter 10 Conic Sections

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

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

In this example, the centre is at $(0, 0)$, so $h = 0$ and $k = 0$. The radius is $r$.

Substitute the values of $h$ and $k$ into the standard equation:

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

$x^2 + y^2 = r^2$

The equation of the circle with centre at (0,0) and radius r is $x^2 + y^2 = r^2$.

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

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

In this example, the centre is at $(-3, 2)$, so $h = -3$ and $k = 2$. The radius is $r = 4$.

Substitute the values of $h$, $k$, and $r$ into the standard equation:

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

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

Expand the squared terms:

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

Combine like terms and move the constant to the left side:

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

$x^2 + y^2 + 6x - 4y - 3 = 0$

The equation of the circle with centre (–3, 2) and radius 4 is $(x + 3)^2 + (y - 2)^2 = 16$ or $x^2 + y^2 + 6x - 4y - 3 = 0$.

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

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

Comparing the given equation with the general equation, we have:

$2g = 8 \implies g = 4$

$2f = 10 \implies f = 5$

$c = -8$

The coordinates of the centre of the circle are $(h, k) = (-g, -f)$.

Centre $(h, k) = (-4, -5)$.

The radius of the circle is given by the formula $r = \sqrt{g^2 + f^2 - c}$.

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

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

$r = \sqrt{49}$

$r = 7$

Alternatively, we can find the centre and radius by completing the square.

Rearrange the terms of the equation:

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

Complete the square for the x-terms by adding $\left(\frac{8}{2}\right)^2 = 4^2 = 16$ to both sides.

Complete the square for the y-terms by adding $\left(\frac{10}{2}\right)^2 = 5^2 = 25$ to both sides.

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

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

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

Comparing $(x + 4)^2 + (y + 5)^2 = 49$ with the standard form:

$x - h = x + 4 \implies h = -4$

$y - k = y + 5 \implies k = -5$

$r^2 = 49 \implies r = \sqrt{49} = 7$ (since the radius must be positive)

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

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

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

The centre $(h, k)$ lies on the line $x + y = 2$. So, its coordinates must satisfy the equation of the line:

The circle passes through the point $(2, -2)$. The distance from the centre $(h, k)$ to this point is the radius $r$.

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

The circle also passes through the point $(3, 4)$. The distance from the centre $(h, k)$ to this point is also the radius $r$.

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

Equating equations (ii) and (iii) since both are equal to $r^2$:

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

Expand the squared terms:

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

Simplify the equation by cancelling $h^2$ and $k^2$ from both sides:

$4 - 4h + 4 + 4k = 9 - 6h + 16 - 8k$

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

Group the terms with $h$ and $k$:

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

Now we have a system of two linear equations with two variables $h$ and $k$ (equations (i) and (iv)):

$h + k = 2 \implies k = 2 - h$ (from (i))

$2h + 12k = 17$ (from (iv))

Substitute $k = 2 - h$ from equation (i) into equation (iv):

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

$2h + 24 - 12h = 17$

$-10h = 17 - 24$

$-10h = -7$

$h = \frac{-7}{-10} = \frac{7}{10}$

Substitute the value of $h$ back into $k = 2 - h$ to find $k$:

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

The centre of the circle is $(h, k) = \left(\frac{7}{10}, \frac{13}{10}\right)$.

Now, substitute the values of $h$ and $k$ into equation (ii) (or (iii)) to find the radius squared $r^2$. Using equation (ii) and point $(2, -2)$:

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

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

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

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

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

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

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

$\left(x - \frac{7}{10}\right)^2 + \left(y - \frac{13}{10}\right)^2 = \frac{1258}{100}$

To write the equation without fractions, we can expand and multiply by 100, or rewrite the left side using a common denominator first:

$\left(\frac{10x - 7}{10}\right)^2 + \left(\frac{10y - 13}{10}\right)^2 = \frac{1258}{100}$

$\frac{(10x - 7)^2}{100} + \frac{(10y - 13)^2}{100} = \frac{1258}{100}$

Multiply both sides by 100:

$(10x - 7)^2 + (10y - 13)^2 = 1258$

Expand the squared terms:

$(100x^2 - 140x + 49) + (100y^2 - 260y + 169) = 1258$

Rearrange and combine constant terms:

$100x^2 + 100y^2 - 140x - 260y + 49 + 169 - 1258 = 0$

$100x^2 + 100y^2 - 140x - 260y - 1040 = 0$

Divide the entire equation by 20 to simplify:

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

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

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

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

Given centre $(h, k) = (0, 2)$ and radius $r = 2$.

Substitute the values into the equation:

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

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

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

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

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

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

The equation of the circle with centre (0, 2) and radius 2 is $x^2 + (y - 2)^2 = 4$ or $x^2 + y^2 - 4y = 0$.

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

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

Given centre $(h, k) = (-2, 3)$ and radius $r = 4$.

Substitute the values into the equation:

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

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

Expand the squared terms:

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

Combine like terms and move the constant to the left side:

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

$x^2 + y^2 + 4x - 6y - 3 = 0$

The equation of the circle with centre (–2, 3) and radius 4 is $(x + 2)^2 + (y - 3)^2 = 16$ or $x^2 + y^2 + 4x - 6y - 3 = 0$.

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

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

Given centre $(h, k) = \left(\frac{1}{2}, \frac{1}{4}\right)$ and radius $r = \frac{1}{12}$.

Substitute the values into the equation:

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

$\left(x - \frac{1}{2}\right)^2 + \left(y - \frac{1}{4}\right)^2 = \frac{1}{144}$

Expand the squared terms:

$\left(x^2 - 2x\left(\frac{1}{2}\right) + \left(\frac{1}{2}\right)^2\right) + \left(y^2 - 2y\left(\frac{1}{4}\right) + \left(\frac{1}{4}\right)^2\right) = \frac{1}{144}$

$\left(x^2 - x + \frac{1}{4}\right) + \left(y^2 - \frac{1}{2}y + \frac{1}{16}\right) = \frac{1}{144}$}

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

To eliminate the fractions, multiply the entire equation by the LCM of 4, 16, and 144, which is 144:

$144 \left(x^2 - x + \frac{1}{4} + y^2 - \frac{1}{2}y + \frac{1}{16}\right) = 144 \left(\frac{1}{144}\right)$

$144x^2 - 144x + 144\left(\frac{1}{4}\right) + 144y^2 - 144\left(\frac{1}{2}\right)y + 144\left(\frac{1}{16}\right) = 1$

$144x^2 - 144x + 36 + 144y^2 - 72y + 9 = 1$

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

$144x^2 + 144y^2 - 144x - 72y + 36 + 9 - 1 = 0$

$144x^2 + 144y^2 - 144x - 72y + 44 = 0$

Divide by 4 to simplify:

$36x^2 + 36y^2 - 36x - 18y + 11 = 0$

The equation of the circle is $\left(x - \frac{1}{2}\right)^2 + \left(y - \frac{1}{4}\right)^2 = \frac{1}{144}$ or $36x^2 + 36y^2 - 36x - 18y + 11 = 0$.

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

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

Given centre $(h, k) = (1, 1)$ and radius $r = \sqrt{2}$.

Substitute the values into the equation:

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

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

Expand the squared terms:

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

Combine like terms and move the constant to the left side:

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

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

The equation of the circle with centre (1, 1) and radius $\sqrt{2}$ is $(x - 1)^2 + (y - 1)^2 = 2$ or $x^2 + y^2 - 2x - 2y = 0$.

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

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

Given centre $(h, k) = (-a, -b)$ and radius $r = \sqrt{a^2 - b^2}$.

For the radius to be a real number, we must have $a^2 - b^2 \geq 0$, which means $a^2 \geq b^2$.

Substitute the values into the equation:

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

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

Expand the squared terms:

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

Combine like terms and move terms to one side:

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

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

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

The equation of the circle with centre (–a, –b) and radius $\sqrt{a^2 - b^2}$ is $(x + a)^2 + (y + b)^2 = a^2 - b^2$ or $x^2 + y^2 + 2ax + 2by + 2b^2 = 0$ (assuming $a^2 \geq b^2$).

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

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

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

Compare the given equation with the standard equation:

$(x + 5)^2 = (x - (-5))^2 \implies h = -5$

$(y - 3)^2 \implies k = 3$

$r^2 = 36 \implies r = \sqrt{36} = 6$ (since the radius is positive)

The centre of the circle is $(-5, 3)$ and the radius is $6$ units.

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

Method 1: Using the general form

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

Comparing the given equation with the general equation, we have:

$2g = -4 \implies g = -2$

$2f = -8 \implies f = -4$

$c = -45$

The coordinates of the centre of the circle are $(h, k) = (-g, -f)$.

Centre $(h, k) = (-(-2), -(-4)) = (2, 4)$.

The radius of the circle is given by the formula $r = \sqrt{g^2 + f^2 - c}$.

$r = \sqrt{(-2)^2 + (-4)^2 - (-45)}$

$r = \sqrt{4 + 16 + 45}$

$r = \sqrt{65}$

Method 2: Completing the square

Rearrange the terms of the equation:

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

Complete the square for the x-terms by adding $\left(\frac{-4}{2}\right)^2 = (-2)^2 = 4$ to both sides.

Complete the square for the y-terms by adding $\left(\frac{-8}{2}\right)^2 = (-4)^2 = 16$ to both sides.

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

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

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

Comparing $(x - 2)^2 + (y - 4)^2 = 65$ with the standard form:

$x - h = x - 2 \implies h = 2$

$y - k = y - 4 \implies k = 4$

$r^2 = 65 \implies r = \sqrt{65}$ (since the radius must be positive)

The centre of the circle is $(2, 4)$ and the radius is $\sqrt{65}$ units.

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

Method 1: Using the general form

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

Comparing the given equation with the general equation, we have:

$2g = -8 \implies g = -4$

$2f = 10 \implies f = 5$

$c = -12$

The coordinates of the centre of the circle are $(h, k) = (-g, -f)$.

Centre $(h, k) = (-(-4), -5) = (4, -5)$.

The radius of the circle is given by the formula $r = \sqrt{g^2 + f^2 - c}$.

$r = \sqrt{(-4)^2 + 5^2 - (-12)}$

$r = \sqrt{16 + 25 + 12}$

$r = \sqrt{53}$

Method 2: Completing the square

Rearrange the terms of the equation:

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

Complete the square for the x-terms by adding $\left(\frac{-8}{2}\right)^2 = (-4)^2 = 16$ to both sides.

Complete the square for the y-terms by adding $\left(\frac{10}{2}\right)^2 = 5^2 = 25$ to both sides.

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

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

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

Comparing $(x - 4)^2 + (y + 5)^2 = 53$ with the standard form:

$x - h = x - 4 \implies h = 4$

$y - k = y + 5 \implies k = -5$

$r^2 = 53 \implies r = \sqrt{53}$ (since the radius must be positive)

The centre of the circle is $(4, -5)$ and the radius is $\sqrt{53}$ units.

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

To find the centre and radius, we need to rewrite the equation in the general form $x^2 + y^2 + 2gx + 2fy + c = 0$, where the coefficients of $x^2$ and $y^2$ are 1.

Divide the entire equation by 2:

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

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

Method 1: Using the general form

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

Comparing the equation $x^2 + y^2 - \frac{1}{2}x = 0$ with the general form, we have:

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

$2f = 0$ (since there is no y-term) $\implies f = 0$

$c = 0$ (since there is no constant term)

The coordinates of the centre of the circle are $(h, k) = (-g, -f)$.

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

The radius of the circle is given by the formula $r = \sqrt{g^2 + f^2 - c}$.

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

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

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

$r = \frac{1}{4}$ (since the radius must be positive)

Method 2: Completing the square

Rearrange the terms of the equation:

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

Complete the square for the x-terms by adding $\left(\frac{-\frac{1}{2}}{2}\right)^2 = \left(-\frac{1}{4}\right)^2 = \frac{1}{16}$ to both sides.

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

$\left(x - \frac{1}{4}\right)^2 + (y - 0)^2 = \frac{1}{16}$}

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

Comparing $\left(x - \frac{1}{4}\right)^2 + (y - 0)^2 = \frac{1}{16}$ with the standard form:

$x - h = x - \frac{1}{4} \implies h = \frac{1}{4}$

$y - k = y - 0 \implies k = 0$

$r^2 = \frac{1}{16} \implies r = \sqrt{\frac{1}{16}} = \frac{1}{4}$ (since the radius must be positive)

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

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

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

The centre $(h, k)$ lies on the line $4x + y = 16$. So, its coordinates must satisfy the equation of the line:

The circle passes through the point $(4, 1)$. The distance from the centre $(h, k)$ to this point is the radius $r$.

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

The circle also passes through the point $(6, 5)$. The distance from the centre $(h, k)$ to this point is also the radius $r$.

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

Equating equations (ii) and (iii) since both are equal to $r^2$:

$(4 - h)^2 + (1 - k)^2 = (6 - h)^2 + (5 - k)^2$

Expand the squared terms:

$(16 - 8h + h^2) + (1 - 2k + k^2) = (36 - 12h + h^2) + (25 - 10k + k^2)$

Simplify the equation by cancelling $h^2$ and $k^2$ from both sides:

$16 - 8h + 1 - 2k = 36 - 12h + 25 - 10k$

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

Group the terms with $h$ and $k$:

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

Divide equation (iv) by 4 to simplify:

Now we have a system of two linear equations with two variables $h$ and $k$ (equations (i) and (v)):

From equation (i), we can express $k$ in terms of $h$: $k = 16 - 4h$.

Substitute this into equation (v):

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

$h + 32 - 8h = 11$

$-7h = 11 - 32$

$-7h = -21$

$h = \frac{-21}{-7} = 3$}

Substitute the value of $h$ back into $k = 16 - 4h$ to find $k$:

$k = 16 - 4(3) = 16 - 12 = 4$

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

Now, substitute the values of $h$ and $k$ into equation (ii) (or (iii)) to find the radius squared $r^2$. Using equation (ii) and point $(4, 1)$:

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

$r^2 = (4 - 3)^2 + (1 - 4)^2$

$r^2 = (1)^2 + (-3)^2$

$r^2 = 1 + 9 = 10$

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

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

$(x - 3)^2 + (y - 4)^2 = 10$

Expand the squared terms to get the general form:

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

$x^2 + y^2 - 6x - 8y + 9 + 16 - 10 = 0$

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

The equation of the circle is $(x - 3)^2 + (y - 4)^2 = 10$ or $x^2 + y^2 - 6x - 8y + 15 = 0$.

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

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

The centre $(h, k)$ lies on the line $x - 3y - 11 = 0$. So, its coordinates must satisfy the equation of the line:

The circle passes through the point $(2, 3)$. The distance from the centre $(h, k)$ to this point is the radius $r$.

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

The circle also passes through the point $(-1, 1)$. The distance from the centre $(h, k)$ to this point is also the radius $r$.

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

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

Equating equations (ii) and (iii) since both are equal to $r^2$:

$(2 - h)^2 + (3 - k)^2 = (1 + h)^2 + (1 - k)^2$}

Expand the squared terms:

$(4 - 4h + h^2) + (9 - 6k + k^2) = (1 + 2h + h^2) + (1 - 2k + k^2)$

Simplify the equation by cancelling $h^2$ and $k^2$ from both sides:

$4 - 4h + 9 - 6k = 1 + 2h + 1 - 2k$

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

Group the terms with $h$ and $k$:

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

Now we have a system of two linear equations with two variables $h$ and $k$ (equations (i) and (iv)):

From equation (i), we can express $h$ in terms of $k$: $h = 11 + 3k$.

Substitute this into equation (iv):

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

$66 + 18k + 4k = 11$

$66 + 22k = 11$

$22k = 11 - 66$

$22k = -55$}

$k = \frac{-55}{22} = -\frac{5}{2}$

Substitute the value of $k$ back into $h = 11 + 3k$ to find $h$:

$h = 11 + 3\left(-\frac{5}{2}\right) = 11 - \frac{15}{2} = \frac{22 - 15}{2} = \frac{7}{2}$

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

Now, substitute the values of $h$ and $k$ into equation (ii) (or (iii)) to find the radius squared $r^2$. Using equation (ii) and point $(2, 3)$:

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

$r^2 = \left(2 - \frac{7}{2}\right)^2 + \left(3 - \left(-\frac{5}{2}\right)\right)^2$

$r^2 = \left(\frac{4 - 7}{2}\right)^2 + \left(3 + \frac{5}{2}\right)^2$

$r^2 = \left(-\frac{3}{2}\right)^2 + \left(\frac{6 + 5}{2}\right)^2$

$r^2 = \left(-\frac{3}{2}\right)^2 + \left(\frac{11}{2}\right)^2$

$r^2 = \frac{9}{4} + \frac{121}{4}$

$r^2 = \frac{9 + 121}{4} = \frac{130}{4} = \frac{65}{2}$

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

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

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

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

To write the equation without fractions, we can expand and multiply by 4, or rewrite the left side using a common denominator first:

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

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

Multiply both sides by 4:

$(2x - 7)^2 + (2y + 5)^2 = 4 \times \frac{65}{2}$

$(2x - 7)^2 + (2y + 5)^2 = 2 \times 65$

$(2x - 7)^2 + (2y + 5)^2 = 130$

Expand the squared terms to get the general form:

$(4x^2 - 28x + 49) + (4y^2 + 20y + 25) = 130$

$4x^2 + 4y^2 - 28x + 20y + 49 + 25 - 130 = 0$

$4x^2 + 4y^2 - 28x + 20y + 74 - 130 = 0$

$4x^2 + 4y^2 - 28x + 20y - 56 = 0$

Divide the entire equation by 4 to simplify:

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

The equation of the circle is $\left(x - \frac{7}{2}\right)^2 + \left(y + \frac{5}{2}\right)^2 = \frac{65}{2}$ or $x^2 + y^2 - 7x + 5y - 14 = 0$.

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

The centre of the circle lies on the x-axis. Therefore, the y-coordinate of the centre is 0. Let the centre be $(h, 0)$.

The equation of the circle with centre $(h, k) = (h, 0)$ and radius $r = 5$ is:

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

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

The circle passes through the point $(2, 3)$. This means the coordinates of this point must satisfy the equation of the circle.

Substitute $x = 2$ and $y = 3$ into the equation:

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

$(2 - h)^2 + 9 = 25$

$(2 - h)^2 = 25 - 9$

$(2 - h)^2 = 16$

Take the square root of both sides:

$2 - h = \pm \sqrt{16}$

$2 - h = \pm 4$

This gives two possible values for $2 - h$:

Case 1: $2 - h = 4$

$-h = 4 - 2$

$-h = 2 \implies h = -2$

The centre is $(-2, 0)$.

The equation of the circle is $(x - (-2))^2 + (y - 0)^2 = 25$

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

Case 2: $2 - h = -4$

$-h = -4 - 2$

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

The centre is $(6, 0)$.

The equation of the circle is $(x - 6)^2 + (y - 0)^2 = 25$

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

Expand the equations to the general form:

Case 1: $(x + 2)^2 + y^2 = 25 \implies x^2 + 4x + 4 + y^2 = 25 \implies x^2 + y^2 + 4x - 21 = 0$

Case 2: $(x - 6)^2 + y^2 = 25 \implies x^2 - 12x + 36 + y^2 = 25 \implies x^2 + y^2 - 12x + 11 = 0$

There are two such circles. Their equations are $(x + 2)^2 + y^2 = 25$ and $(x - 6)^2 + y^2 = 25$ (or in general form: $x^2 + y^2 + 4x - 21 = 0$ and $x^2 + y^2 - 12x + 11 = 0$).

The circle passes through the origin $(0, 0)$.

The circle makes an intercept $a$ on the x-axis. This means the circle intersects the x-axis at two points. One point is the origin $(0, 0)$. Let the other point be $(a, 0)$.

The circle makes an intercept $b$ on the y-axis. This means the circle intersects the y-axis at two points. One point is the origin $(0, 0)$. Let the other point be $(0, b)$.

Since the circle passes through $(0, 0)$, $(a, 0)$, and $(0, b)$, these three points lie on the circle.

Let the general equation of the circle be $x^2 + y^2 + 2gx + 2fy + c = 0$.

The circle passes through $(0, 0)$: Substitute $x=0, y=0$ into the equation:

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

$c = 0$

The circle passes through $(a, 0)$: Substitute $x=a, y=0$ and $c=0$ into the equation:

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

$a^2 + 2ga = 0$

$a(a + 2g) = 0$

This gives $a = 0$ or $a + 2g = 0$. If the x-intercept is non-zero ($a \neq 0$), then $a + 2g = 0 \implies 2g = -a \implies g = -\frac{a}{2}$.

The circle passes through $(0, b)$: Substitute $x=0, y=b$ and $c=0$ into the equation:

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

$b^2 + 2fb = 0$

$b(b + 2f) = 0$

This gives $b = 0$ or $b + 2f = 0$. If the y-intercept is non-zero ($b \neq 0$), then $b + 2f = 0 \implies 2f = -b \implies f = -\frac{b}{2}$.

Substitute the values of $g$, $f$, and $c$ back into the general equation $x^2 + y^2 + 2gx + 2fy + c = 0$:

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

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

This derivation assumes $a \neq 0$ and $b \neq 0$.

If $a = 0$, the circle passes through $(0, 0)$ and $(0, 0)$, which doesn't give a distinct intercept point on the x-axis other than the origin unless the circle is just the point (0,0). Similarly for $b=0$. The question implies distinct intercepts, so $a \neq 0$ and $b \neq 0$.

The equation of the circle passing through (0, 0) and making intercepts a and b on the coordinate axes is $x^2 + y^2 - ax - by = 0$.

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

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

The radius $r$ of the circle is the distance between the centre $(2, 2)$ and the point $(4, 5)$.

Using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$:

$r = \sqrt{(4 - 2)^2 + (5 - 2)^2}$

$r = \sqrt{(2)^2 + (3)^2}$

$r = \sqrt{4 + 9}$

$r = \sqrt{13}$

The 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 $(h, k) = (2, 2)$ and radius $r = \sqrt{13}$ into the equation:

$(x - 2)^2 + (y - 2)^2 = (\sqrt{13})^2$

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

Expand the squared terms to get the general form:

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

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

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

The equation of the circle with centre (2, 2) and passing through the point (4, 5) is $(x - 2)^2 + (y - 2)^2 = 13$ or $x^2 + y^2 - 4x - 4y - 5 = 0$.

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

This is the equation of a circle with centre at the origin $(0, 0)$ and radius $r^2 = 25$, so $r = \sqrt{25} = 5$.

To determine if a point $(x_0, y_0)$ lies inside, outside, or on the circle $x^2 + y^2 = r^2$, we compare the square of the distance of the point from the centre $(0, 0)$ with the square of the radius $r^2$.

The square of the distance of the point $(x_0, y_0)$ from the centre $(0, 0)$ is $d^2 = (x_0 - 0)^2 + (y_0 - 0)^2 = x_0^2 + y_0^2$.

• If $x_0^2 + y_0^2 < r^2$, the point lies inside the circle.
• If $x_0^2 + y_0^2 = r^2$, the point lies on the circle.
• If $x_0^2 + y_0^2 > r^2$, the point lies outside the circle.

The given point is $(-2.5, 3.5)$. So, $x_0 = -2.5$ and $y_0 = 3.5$.

Calculate $x_0^2 + y_0^2$:

$x_0^2 = (-2.5)^2 = (-5/2)^2 = \frac{25}{4} = 6.25$

$y_0^2 = (3.5)^2 = (7/2)^2 = \frac{49}{4} = 12.25$

$x_0^2 + y_0^2 = 6.25 + 12.25 = 18.50$

The square of the radius is $r^2 = 25$.

Compare $x_0^2 + y_0^2$ with $r^2$:

$18.50$ vs $25$

Since $18.50 < 25$, the square of the distance of the point from the centre is less than the square of the radius.

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

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

This equation is in the standard form $y^2 = 4ax$.

Comparing $y^2 = 8x$ with $y^2 = 4ax$, we have:

$4a = 8$

$a = \frac{8}{4} = 2$

For a parabola of the form $y^2 = 4ax$ where $a > 0$:

Focus: The coordinates of the focus are $(a, 0)$.

Focus: $(2, 0)$.

Axis: The axis of the parabola is the x-axis (the axis of symmetry). The equation of the x-axis is $y = 0$.

Axis: $y = 0$.

Directrix: The equation of the directrix is $x = -a$.

Directrix: $x = -2$. This can also be written as $x + 2 = 0$.

Latus Rectum: The length of the latus rectum is $4a$.

Length of Latus Rectum: $4(2) = 8$.

The equation of the latus rectum is a vertical line through the focus $(a, 0)$, i.e., $x = a$.

Equation of Latus Rectum: $x = 2$. This can also be written as $x - 2 = 0$.

Summary:

Coordinates of the focus: $(2, 0)$

Equation of the axis: $y = 0$

Equation of the directrix: $x = -2$

Length of the latus rectum: $8$ units

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

The equation of the directrix is given as $x = -2$, which can be written as $x + 2 = 0$.

By definition, a parabola is the locus of a point P(x, y) such that its distance from the focus is equal to its perpendicular distance from the directrix.

Let P(x, y) be a point on the parabola.

Distance from P(x, y) to the focus F(2, 0):

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

The perpendicular distance from P(x, y) to the directrix $x + 2 = 0$ is given by the formula $\frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$.

Here, $(x_0, y_0) = (x, y)$, and for the line $x + 2 = 0$, we have $A=1$, $B=0$, $C=2$.

Distance from P(x, y) to the directrix: $PD = \frac{|1(x) + 0(y) + 2|}{\sqrt{1^2 + 0^2}} = \frac{|x + 2|}{\sqrt{1}} = |x + 2|$

According to the definition of a parabola, $PF = PD$:

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

Square both sides of the equation to eliminate the square root and the absolute value:

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

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

Expand both sides:

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

Simplify the equation:

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

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

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

Move the -4x term to the right side:

$y^2 = 4x + 4x$

$y^2 = 8x$

Alternatively, since the focus is at $(2, 0)$ and the directrix is $x = -2$, the axis of the parabola is the x-axis and the vertex is at the midpoint of the segment joining the focus and the point on the directrix closest to the focus, which is $(-2, 0)$. The vertex is $\left(\frac{2 + (-2)}{2}, \frac{0 + 0}{2}\right) = (0, 0)$.

The distance from the vertex to the focus is $a = 2$. Since the focus is on the positive x-axis, the parabola opens to the right. The standard equation for such a parabola is $y^2 = 4ax$.

Substitute $a = 2$ into the standard equation:

$y^2 = 4(2)x$

$y^2 = 8x$

The equation of the parabola with focus (2, 0) and directrix x = – 2 is $y^2 = 8x$.

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

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

Since the vertex is at $(0, 0)$ and the focus is at $(0, 2)$, which is on the positive y-axis, the axis of the parabola is the y-axis and the parabola opens upwards.

The standard equation of a parabola with vertex at $(0, 0)$ and opening upwards is $x^2 = 4ay$, where $a$ is the distance from the vertex to the focus.

The distance from the vertex $(0, 0)$ to the focus $(0, 2)$ is:

$a = \sqrt{(0 - 0)^2 + (2 - 0)^2} = \sqrt{0^2 + 2^2} = \sqrt{4} = 2$

Since the focus is on the positive y-axis, the value of $a$ for the standard form $x^2 = 4ay$ is 2.

Substitute $a = 2$ into the standard equation $x^2 = 4ay$:

$x^2 = 4(2)y$

$x^2 = 8y$

The equation of the parabola with vertex at (0, 0) and focus at (0, 2) is $x^2 = 8y$.

The parabola is symmetric about the y-axis. This means its axis of symmetry is the y-axis.

The standard equations for parabolas symmetric about the y-axis with vertex at the origin are $x^2 = 4ay$ (opens upwards) or $x^2 = -4ay$ (opens downwards).

We can write the general form for a parabola symmetric about the y-axis with vertex at the origin as $x^2 = ky$, where $k$ is a constant ($k = 4a$ or $k = -4a$).

The parabola passes through the point $(2, -3)$. This means the coordinates of this point must satisfy the equation of the parabola.

Substitute $x = 2$ and $y = -3$ into the equation $x^2 = ky$:

$(2)^2 = k(-3)$

$4 = -3k$

Solving for $k$:

$k = -\frac{4}{3}$

Substitute the value of $k$ back into the equation $x^2 = ky$:

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

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

We can also write the equation by multiplying by 3:

$3x^2 = -4y$

or

$3x^2 + 4y = 0$

Since $k = -\frac{4}{3}$ is negative, the parabola opens downwards, which is consistent with it passing through the point $(2, -3)$ which is in the fourth quadrant.

The equation of the parabola which is symmetric about the y-axis and passes through the point (2, –3) is $x^2 = -\frac{4}{3}y$ or $3x^2 + 4y = 0$.

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

This equation is in the standard form $y^2 = 4ax$.

Comparing $y^2 = 12x$ with $y^2 = 4ax$, we have:

$4a = 12$

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

For a parabola of the form $y^2 = 4ax$ where $a > 0$:

Focus: The coordinates of the focus are $(a, 0)$.

Focus: $(3, 0)$.

Axis: The axis of the parabola is the x-axis (the axis of symmetry). The equation of the x-axis is $y = 0$.

Axis: $y = 0$.

Directrix: The equation of the directrix is $x = -a$.

Directrix: $x = -3$. This can also be written as $x + 3 = 0$.

Latus Rectum: The length of the latus rectum is $4a$.

Length of Latus Rectum: $4(3) = 12$.

Summary:

Coordinates of the focus: $(3, 0)$

Equation of the axis: $y = 0$

Equation of the directrix: $x = -3$

Length of the latus rectum: $12$ units

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

This equation is in the standard form $x^2 = 4ay$.

Comparing $x^2 = 6y$ with $x^2 = 4ay$, we have:

$4a = 6$

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

For a parabola of the form $x^2 = 4ay$ where $a > 0$ (it opens upwards):

Focus: The coordinates of the focus are $(0, a)$.

Focus: $\left(0, \frac{3}{2}\right)$.

Axis: The axis of the parabola is the y-axis (the axis of symmetry). The equation of the y-axis is $x = 0$.

Axis: $x = 0$.

Directrix: The equation of the directrix is $y = -a$.

Directrix: $y = -\frac{3}{2}$. This can also be written as $2y = -3$ or $2y + 3 = 0$.

Latus Rectum: The length of the latus rectum is $4a$.

Length of Latus Rectum: $4\left(\frac{3}{2}\right) = 2 \times 3 = 6$.

Summary:

Coordinates of the focus: $\left(0, \frac{3}{2}\right)$

Equation of the axis: $x = 0$

Equation of the directrix: $y = -\frac{3}{2}$

Length of the latus rectum: $6$ units

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

This equation is in the standard form $y^2 = -4ax$.

Comparing $y^2 = -8x$ with $y^2 = -4ax$, we have:

$4a = 8$

$a = \frac{8}{4} = 2$

For a parabola of the form $y^2 = -4ax$ where $a > 0$ (it opens to the left):

Focus: The coordinates of the focus are $(-a, 0)$.

Focus: $(-2, 0)$.

Axis: The axis of the parabola is the x-axis (the axis of symmetry). The equation of the x-axis is $y = 0$.

Axis: $y = 0$.

Directrix: The equation of the directrix is $x = a$.

Directrix: $x = 2$. This can also be written as $x - 2 = 0$.

Latus Rectum: The length of the latus rectum is $4a$.

Length of Latus Rectum: $4(2) = 8$.

Summary:

Coordinates of the focus: $(-2, 0)$

Equation of the axis: $y = 0$

Equation of the directrix: $x = 2$

Length of the latus rectum: $8$ units

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

This equation is in the standard form $x^2 = -4ay$.

Comparing $x^2 = -16y$ with $x^2 = -4ay$, we have:

$4a = 16$

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

For a parabola of the form $x^2 = -4ay$ where $a > 0$ (it opens downwards):

Focus: The coordinates of the focus are $(0, -a)$.

Focus: $(0, -4)$.

Axis: The axis of the parabola is the y-axis (the axis of symmetry). The equation of the y-axis is $x = 0$.

Axis: $x = 0$.

Directrix: The equation of the directrix is $y = a$.

Directrix: $y = 4$. This can also be written as $y - 4 = 0$.

Latus Rectum: The length of the latus rectum is $4a$.

Length of Latus Rectum: $4(4) = 16$.

Summary:

Coordinates of the focus: $(0, -4)$

Equation of the axis: $x = 0$

Equation of the directrix: $y = 4$

Length of the latus rectum: $16$ units

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

This equation is in the standard form $y^2 = 4ax$.

Comparing $y^2 = 10x$ with $y^2 = 4ax$, we have:

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

For a parabola of the form $y^2 = 4ax$ where $a > 0$ (it opens to the right):

Focus: The coordinates of the focus are $(a, 0)$.

Focus: $\left(\frac{5}{2}, 0\right)$.

Axis: The axis of the parabola is the x-axis (the axis of symmetry). The equation of the x-axis is $y = 0$.

Axis: $y = 0$.

Directrix: The equation of the directrix is $x = -a$.

Directrix: $x = -\frac{5}{2}$. This can also be written as $2x = -5$ or $2x + 5 = 0$.

Latus Rectum: The length of the latus rectum is $4a$.

Length of Latus Rectum: $4\left(\frac{5}{2}\right) = 2 \times 5 = 10$.

Summary:

Coordinates of the focus: $\left(\frac{5}{2}, 0\right)$

Equation of the axis: $y = 0$

Equation of the directrix: $x = -\frac{5}{2}$

Length of the latus rectum: $10$ units

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

This equation is in the standard form $x^2 = -4ay$.

Comparing $x^2 = -9y$ with $x^2 = -4ay$, we have:

$4a = 9$

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

For a parabola of the form $x^2 = -4ay$ where $a > 0$ (it opens downwards):

Focus: The coordinates of the focus are $(0, -a)$.

Focus: $\left(0, -\frac{9}{4}\right)$.

Axis: The axis of the parabola is the y-axis (the axis of symmetry). The equation of the y-axis is $x = 0$.

Axis: $x = 0$.

Directrix: The equation of the directrix is $y = a$.

Directrix: $y = \frac{9}{4}$. This can also be written as $4y = 9$ or $4y - 9 = 0$.

Latus Rectum: The length of the latus rectum is $4a$.

Length of Latus Rectum: $4\left(\frac{9}{4}\right) = 9$.

Summary:

Coordinates of the focus: $\left(0, -\frac{9}{4}\right)$

Equation of the axis: $x = 0$

Equation of the directrix: $y = \frac{9}{4}$

Length of the latus rectum: $9$ units

Given:

Focus of the parabola $F = (6, 0)$.

Equation of the directrix is $x = -6$, which can be written as $x + 6 = 0$.

To Find:

The equation of the parabola.

Solution 1 (Using Definition of Parabola):

A parabola is defined as the locus of a point that moves in a plane such that its distance from a fixed point (the focus) is equal to its distance from a fixed straight line (the directrix).

Let $P(x, y)$ be any point on the parabola.

The distance of point $P(x, y)$ from the focus $F(6, 0)$ is given by the distance formula:

$PF = \sqrt{(x - 6)^2 + (y - 0)^2} = \sqrt{(x - 6)^2 + y^2}$.

The distance of point $P(x, y)$ from the directrix $x + 6 = 0$ is given by the formula for the perpendicular distance from a point to a line:

$PL = \frac{|x + 6|}{\sqrt{1^2 + 0^2}} = |x + 6|$.

By the definition of a parabola, the distance $PF$ must be equal to the distance $PL$ for any point $P(x, y)$ on the parabola.

$PF = PL$

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

To eliminate the square root and absolute value, we square both sides of the equation:

Now, we expand both sides of equation (i):

Expanding the left side:

$(x - 6)^2 + y^2 = (x^2 - 2 \cdot x \cdot 6 + 6^2) + y^2 = x^2 - 12x + 36 + y^2$.

Expanding the right side:

$(x + 6)^2 = x^2 + 2 \cdot x \cdot 6 + 6^2 = x^2 + 12x + 36$.

Substitute these expanded forms back into equation (i):

Now, we simplify the equation. Subtract $x^2$ and $36$ from both sides:

Add $12x$ to both sides of the equation:

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

Solution 2 (Using Standard Form):

The standard equation of a parabola with its vertex at the origin $(0, 0)$, focus at $(a, 0)$, and directrix as the vertical line $x = -a$ is given by $y^2 = 4ax$.

In the given problem:

The focus is $(6, 0)$. Comparing this with the standard focus $(a, 0)$, we find that $a = 6$.

The directrix is $x = -6$. Comparing this with the standard directrix $x = -a$, we have $x = -6$, which means $-a = -6$, so $a = 6$.

Since the focus is $(a, 0)$ with $a > 0$ and the directrix is $x = -a$, the parabola opens to the right, and its equation is of the form $y^2 = 4ax$.

Substitute the value of $a = 6$ into the standard equation $y^2 = 4ax$:

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

Given:

Focus of the parabola $F = (0, -3)$.

Equation of the directrix is $y = 3$, which can be written as $y - 3 = 0$.

To Find:

The equation of the parabola.

Solution 1 (Using Definition of Parabola):

A parabola is defined as the locus of a point that moves in a plane such that its distance from a fixed point (the focus) is equal to its distance from a fixed straight line (the directrix).

Let $P(x, y)$ be any point on the parabola.

The distance of point $P(x, y)$ from the focus $F(0, -3)$ is given by the distance formula:

$PF = \sqrt{(x - 0)^2 + (y - (-3))^2} = \sqrt{x^2 + (y + 3)^2}$.

The distance of point $P(x, y)$ from the directrix $y - 3 = 0$ is given by the formula for the perpendicular distance from a point to a line:

$PL = \frac{|0 \cdot x + 1 \cdot y - 3|}{\sqrt{0^2 + 1^2}} = \frac{|y - 3|}{\sqrt{1}} = |y - 3|$.

By the definition of a parabola, the distance $PF$ must be equal to the distance $PL$ for any point $P(x, y)$ on the parabola.

$PF = PL$

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

To eliminate the square root and absolute value, we square both sides of the equation:

Now, we expand both sides of equation (i):

Expanding the left side:

$x^2 + (y + 3)^2 = x^2 + (y^2 + 2 \cdot y \cdot 3 + 3^2) = x^2 + y^2 + 6y + 9$.

Expanding the right side:

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

Substitute these expanded forms back into equation (i):

Now, we simplify the equation. Subtract $y^2$ and $9$ from both sides:

Subtract $6y$ from both sides of the equation:

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

Solution 2 (Using Standard Form):

The standard equation of a parabola with its vertex at the origin $(0, 0)$, focus at $(0, -a)$, and directrix as the horizontal line $y = a$ is given by $x^2 = -4ay$.

In the given problem:

The focus is $(0, -3)$. Comparing this with the standard focus $(0, -a)$, we find that $-a = -3$, so $a = 3$.

The directrix is $y = 3$. Comparing this with the standard directrix $y = a$, we have $y = 3$, which means $a = 3$.

Since the focus is $(0, -a)$ with $a > 0$ and the directrix is $y = a$, the parabola opens downwards, and its equation is of the form $x^2 = -4ay$.

Substitute the value of $a = 3$ into the standard equation $x^2 = -4ay$:

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

Given:

Vertex of the parabola: $(0, 0)$.

Focus of the parabola: $(3, 0)$.

To Find:

The equation of the parabola.

Solution:

The vertex of the parabola is given as $(0, 0)$, which is the origin.

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

Since the vertex is at the origin and the focus is on the x-axis at a positive x-coordinate $(3 > 0)$, the axis of the parabola is the positive x-axis. The parabola opens to the right.

The standard equation of a parabola with vertex at the origin $(0, 0)$ and focus at $(a, 0)$ is given by $y^2 = 4ax$.

Comparing the given focus $(3, 0)$ with the standard focus $(a, 0)$, we find the value of $a$:

Now, substitute the value of $a = 3$ into the standard equation $y^2 = 4ax$:

Simplify the equation:

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

Given:

Vertex of the parabola: $(0, 0)$.

Focus of the parabola: $(-2, 0)$.

To Find:

The equation of the parabola.

Solution:

The vertex of the parabola is given as $(0, 0)$, which is the origin.

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

Since the vertex is at the origin and the focus is on the x-axis at a negative x-coordinate $(-2 < 0)$, the axis of the parabola is the negative x-axis. The parabola opens to the left.

The standard equation of a parabola with vertex at the origin $(0, 0)$ and focus at $(-a, 0)$ is given by $y^2 = -4ax$, where $a > 0$.

Comparing the given focus $(-2, 0)$ with the standard focus $(-a, 0)$, we find the value of $a$:

This implies:

Now, substitute the value of $a = 2$ into the standard equation $y^2 = -4ax$:

Simplify the equation:

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

Vertex of the parabola: $(0, 0)$.

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

The axis of the parabola is along the x-axis.

To Find:

The equation of the parabola.

Solution:

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

The axis of the parabola is along the x-axis. This means the standard equation of the parabola is either $y^2 = 4ax$ (if it opens to the right) or $y^2 = -4ax$ (if it opens to the left).

The parabola passes through the point $(2, 3)$. Since the x-coordinate of this point is positive ($2 > 0$), and the axis is the x-axis, the parabola must open towards the positive x-axis (to the right).

Therefore, the standard equation of the parabola is of the form $y^2 = 4ax$, where $a$ is a positive constant representing the distance from the vertex to the focus.

Since the point $(2, 3)$ lies on the parabola, it must satisfy the equation $y^2 = 4ax$.

Substitute the coordinates of the point $(2, 3)$ into the equation, with $x = 2$ and $y = 3$:

Simplify equation (i):

Now, solve for $a$:

Since $a = \frac{9}{8} > 0$, our assumption that the parabola opens to the right is consistent with the point $(2, 3)$ being on the curve.

Substitute the value of $a = \frac{9}{8}$ back into the standard equation $y^2 = 4ax$:

Simplify the equation:

Alternatively, we can write it as:

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

Given:

Vertex of the parabola: $(0, 0)$.

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:

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

The parabola is symmetric with respect to the y-axis. This means the standard equation of the parabola is either $x^2 = 4ay$ (if it opens upwards) or $x^2 = -4ay$ (if it opens downwards), where $a > 0$ is the distance from the vertex to the focus.

The parabola passes through the point $(5, 2)$. This point lies in the first quadrant, where both x and y coordinates are positive ($5 > 0$ and $2 > 0$).

If the parabola opened downwards ($x^2 = -4ay$), then for any point $(x, y)$ on the parabola (except the vertex), $y$ would have to be negative (since $x^2$ is non-negative and $-4a$ is negative). But the point $(5, 2)$ has a positive y-coordinate ($2$). Therefore, the parabola must open upwards.

The standard equation of a parabola with vertex at the origin $(0, 0)$ and symmetric with respect to the positive y-axis (opening upwards) is given by $x^2 = 4ay$, where $a > 0$.

Since the point $(5, 2)$ lies on the parabola, it must satisfy the equation $x^2 = 4ay$.

Substitute the coordinates of the point $(5, 2)$ into the equation, with $x = 5$ and $y = 2$:

Simplify equation (i):

Now, solve for $a$:

Since $a = \frac{25}{8} > 0$, our assumption that the parabola opens upwards is consistent with the point $(5, 2)$ being on the curve.

Substitute the value of $a = \frac{25}{8}$ back into the standard equation $x^2 = 4ay$:

Simplify the equation:

Alternatively, we can write it as:

Thus, 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, the vertices, the length of major axis, the minor axis, the eccentricity and the latus rectum of the given ellipse.

Solution:

The given equation of the ellipse is:

This equation is of the standard form $ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $.

Comparing equation (1) with the standard form, we have:

From (2) and (3), we get $a = \sqrt{25} = 5$ and $b = \sqrt{9} = 3$.

Since $a^2 = 25 > b^2 = 9$, the major axis of the ellipse is along the x-axis.

The coordinates of the foci are $ (\pm c, 0) $ and the coordinates of the vertices are $ (\pm a, 0) $, where $c^2 = a^2 - b^2$.

Let's find the value of $c$:

Substitute the values of $a^2$ and $b^2$ from (2) and (3) into (4):

$c^2 = 25 - 9$

$c^2 = 16$

$c = \sqrt{16} = 4$

Now we can find the required properties:

Coordinates of the foci: $(\pm c, 0) = (\pm 4, 0)$.

The foci are at $ \mathbf{(4, 0)} $ and $ \mathbf{(-4, 0)} $.

Coordinates of the vertices: $(\pm a, 0) = (\pm 5, 0)$.

The vertices are at $ \mathbf{(5, 0)} $ and $ \mathbf{(-5, 0)} $.

Length of major axis: $2a = 2 \times 5 = \mathbf{10}$ units.

Length of minor axis: $2b = 2 \times 3 = \mathbf{6}$ units.

Eccentricity: $e = \frac{c}{a} = \frac{4}{5} = \mathbf{0.8}$.

Length of latus rectum: $ \frac{2b^2}{a} = \frac{2 \times 9}{5} = \frac{18}{5} = \mathbf{3.6}$ units.

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} $ or $ 0.8 $

Length of Latus Rectum: $ \frac{18}{5} $ or $ 3.6 $

Given:

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

To Find:

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

Solution:

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

To convert this into the standard form $ \frac{x^2}{B^2} + \frac{y^2}{A^2} = 1 $ (where $A$ and $B$ represent the semi-major and semi-minor axes lengths, and the major axis is along the y-axis or x-axis), we need to divide both sides of the equation by 36.

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

Simplify the fractions:

$ \frac{\cancel{9}^{1}x^2}{\cancel{36}_{4}} + \frac{\cancel{4}^{1}y^2}{\cancel{36}_{9}} = 1 $

This gives the standard form of the ellipse equation:

This equation is of the standard form $ \frac{x^2}{b^2} + \frac{y^2}{a^2} = 1 $ (since the denominator of the $y^2$ term is larger, indicating the major axis is along the y-axis).

Comparing equation (1) with the standard form, we have:

From (2) and (3), we get $a = \sqrt{9} = 3$ and $b = \sqrt{4} = 2$.

Since $a^2 = 9 > b^2 = 4$, the major axis of the ellipse is along the y-axis.

The coordinates of the foci are $ (0, \pm c) $ and the coordinates of the vertices are $ (0, \pm a) $, where $c^2 = a^2 - b^2$.

Let's find the value of $c$:

Substitute the values of $a^2$ and $b^2$ from (2) and (3) into (4):

$c^2 = 9 - 4$

$c^2 = 5$

$c = \sqrt{5} $

Now we can find the required properties:

Coordinates of the foci: $ (0, \pm c) = \mathbf{(0, \pm \sqrt{5})} $.

The foci are at $ (0, \sqrt{5}) $ and $ (0, -\sqrt{5}) $.

Coordinates of the vertices: $ (0, \pm a) = \mathbf{(0, \pm 3)} $.

The vertices are at $ (0, 3) $ and $ (0, -3) $.

Length of major axis: $2a = 2 \times 3 = \mathbf{6}$ units.

Length of minor axis: $2b = 2 \times 2 = \mathbf{4}$ units.

Eccentricity: $e = \frac{c}{a} = \mathbf{\frac{\sqrt{5}}{3}}$.

Length of latus rectum: $ \frac{2b^2}{a} = \frac{2 \times 4}{3} = \frac{8}{3}$ units.

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} $

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

Since the vertices are $(\pm 13, 0)$ and the foci are $(\pm 5, 0)$, the center of the ellipse is at the origin $ (0, 0) $.

Also, the vertices and foci lie on the x-axis. This indicates that the major axis of the ellipse is along the x-axis.

The standard form of the equation of an ellipse with the center at the origin and major axis along the x-axis is:

where $a$ is the length of the semi-major axis and $b$ is the length of the semi-minor axis.

The coordinates of the vertices of such an ellipse are $(\pm a, 0)$.

Comparing the given vertices $(\pm 13, 0)$ with $(\pm a, 0)$, we get:

So, $a^2 = 13^2 = 169$.

The coordinates of the foci of such an ellipse are $(\pm c, 0)$, where $c$ is the distance from the center to each focus.

Comparing the given foci $(\pm 5, 0)$ with $(\pm c, 0)$, we get:

So, $c^2 = 5^2 = 25$.

We know the relationship between $a$, $b$, and $c$ for an ellipse is $c^2 = a^2 - b^2$.

We can use this relation to find $b^2$:

Substitute the values of $a^2$ and $c^2$ from (2) and (3) into (4):

$b^2 = 169 - 25$

$b^2 = 144$

Now, substitute the values of $a^2 = 169$ and $b^2 = 144$ into the standard equation of the ellipse (1):

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

Given:

Length of the major axis = 20.

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

To Find:

The equation of the ellipse.

Solution:

Since the foci are located at $(0, \pm 5)$, they lie on the y-axis.

This means the center of the ellipse is at the origin $(0, 0)$ and the major axis is along the y-axis.

The standard form of the equation of an ellipse with the center at the origin and major axis along the y-axis is:

where $a$ is the length of the semi-major axis and $b$ is the length of the semi-minor axis.

The coordinates of the foci of such an ellipse are $(0, \pm c)$, where $c$ is the distance from the center to each focus.

Comparing the given foci $(0, \pm 5)$ with $(0, \pm c)$, we get:

So, $c^2 = 5^2 = 25$.

The length of the major axis is given as 20. For an ellipse with the major axis along the y-axis, the length of the major axis is $2a$.

So, we have:

Dividing by 2, we get:

So, $a^2 = 10^2 = 100$.

We use the relationship between $a$, $b$, and $c$ for an ellipse, which is $c^2 = a^2 - b^2$.

We can rearrange this to find $b^2$:

Substitute the values of $a^2 = 100$ (from 3) and $c^2 = 25$ (from 2) into (4):

$b^2 = 100 - 25$

$b^2 = 75$

Now, substitute the values of $a^2 = 100$ and $b^2 = 75$ into the standard equation of the ellipse (1):

Thus, the equation of the ellipse is $ \mathbf{\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 of the ellipse is along the x-axis and the center is implicitly assumed to be at the origin, the standard form of the equation of the ellipse is:

where $a$ is the length of the semi-major axis and $b$ is the length of the semi-minor axis, and $a > b$.

The ellipse passes through the point $ (4, 3) $. Substituting $x=4$ and $y=3$ into equation (1), we get:

This simplifies to:

The ellipse also passes through the point $ (-1, 4) $. Substituting $x=-1$ and $y=4$ into equation (1), we get:

This simplifies to:

We now have a system of two linear equations in terms of $ \frac{1}{a^2} $ and $ \frac{1}{b^2} $.

Let $ X = \frac{1}{a^2} $ and $ Y = \frac{1}{b^2} $.

The system of equations (3) and (5) becomes:

$16X + 9Y = 1$

$X + 16Y = 1$

From the second equation, we can express $X$ in terms of $Y$:

Substitute equation (6) into the first equation:

$16(1 - 16Y) + 9Y = 1$

$16 - 256Y + 9Y = 1$

$16 - 247Y = 1$

$16 - 1 = 247Y$

$15 = 247Y$

Now substitute the value of $Y$ from (7) into equation (6) to find $X$:

$X = 1 - 16 \left(\frac{15}{247}\right) $

$X = 1 - \frac{16 \times 15}{247} $

$X = 1 - \frac{240}{247} $

$X = \frac{247 - 240}{247} $

Now we have $ \frac{1}{a^2} = X = \frac{7}{247} $ and $ \frac{1}{b^2} = Y = \frac{15}{247} $.

From these, we find $ a^2 $ and $ b^2 $:

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

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

Note that $ \frac{247}{7} > \frac{247}{15} $, so $ a^2 > b^2 $, which is consistent with the major axis being along the x-axis.

Substitute the values of $ a^2 $ and $ b^2 $ back into the standard equation of the ellipse (1):

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

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

Multiply both sides by 247:

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

Thus, the equation of the ellipse is $ \mathbf{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 major axis, 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 of the standard form $ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $.

Comparing equation (1) with the standard form, we have:

From (2) and (3), we get $a = \sqrt{36} = 6$ and $b = \sqrt{16} = 4$.

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

The coordinates of the foci are $ (\pm c, 0) $ and the coordinates of the vertices are $ (\pm a, 0) $, where $c^2 = a^2 - b^2$.

Let's find the value of $c$:

Substitute the values of $a^2$ and $b^2$ from (2) and (3) into (4):

$c^2 = 36 - 16$

$c^2 = 20$

$c = \sqrt{20} = \sqrt{4 \times 5} = 2\sqrt{5} $

Now we can find the required properties:

Coordinates of the foci: $(\pm c, 0) = (\pm 2\sqrt{5}, 0)$.

The foci are at $ \mathbf{(2\sqrt{5}, 0)} $ and $ \mathbf{(-2\sqrt{5}, 0)} $.

Coordinates of the vertices: $(\pm a, 0) = (\pm 6, 0)$.

The vertices are at $ \mathbf{(6, 0)} $ and $ \mathbf{(-6, 0)} $.

Length of major axis: $2a = 2 \times 6 = \mathbf{12}$ units.

Length of minor axis: $2b = 2 \times 4 = \mathbf{8}$ units.

Eccentricity: $e = \frac{c}{a} = \frac{2\sqrt{5}}{6} = \mathbf{\frac{\sqrt{5}}{3}}$.

Length of latus rectum: $ \frac{2b^2}{a} = \frac{2 \times 16}{6} = \frac{32}{6} = \mathbf{\frac{16}{3}}$ units.

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 major axis, 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 of the standard form $ \frac{x^2}{b^2} + \frac{y^2}{a^2} = 1 $ (since the denominator of the $y^2$ term is larger, indicating the major axis is along the y-axis).

Comparing equation (1) with the standard form, we have:

From (2) and (3), we get $a = \sqrt{25} = 5$ and $b = \sqrt{4} = 2$.

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

The coordinates of the foci are $ (0, \pm c) $ and the coordinates of the vertices are $ (0, \pm a) $, where $c^2 = a^2 - b^2$.

Let's find the value of $c$:

Substitute the values of $a^2$ and $b^2$ from (2) and (3) into (4):

$c^2 = 25 - 4$

$c^2 = 21$

$c = \sqrt{21} $

Now we can find the required properties:

Coordinates of the foci: $ (0, \pm c) = \mathbf{(0, \pm \sqrt{21})} $.

The foci are at $ (0, \sqrt{21}) $ and $ (0, -\sqrt{21}) $.

Coordinates of the vertices: $ (0, \pm a) = \mathbf{(0, \pm 5)} $.

The vertices are at $ (0, 5) $ and $ (0, -5) $.

Length of major axis: $2a = 2 \times 5 = \mathbf{10}$ units.

Length of minor axis: $2b = 2 \times 2 = \mathbf{4}$ units.

Eccentricity: $e = \frac{c}{a} = \frac{\sqrt{21}}{5} = \mathbf{\frac{\sqrt{21}}{5}}$.

Length of latus rectum: $ \frac{2b^2}{a} = \frac{2 \times 4}{5} = \frac{8}{5} = \mathbf{1.6}$ units.

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} $ or $ 1.6 $

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 major axis, 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 of the standard form $ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $.

Comparing equation (1) with the standard form, we have:

From (2) and (3), we get $a = \sqrt{16} = 4$ and $b = \sqrt{9} = 3$.

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

The coordinates of the foci are $ (\pm c, 0) $ and the coordinates of the vertices are $ (\pm a, 0) $, where $c^2 = a^2 - b^2$.

Let's find the value of $c$:

Substitute the values of $a^2$ and $b^2$ from (2) and (3) into (4):

$c^2 = 16 - 9$

$c^2 = 7$

$c = \sqrt{7} $

Now we can find the required properties:

Coordinates of the foci: $(\pm c, 0) = (\pm \sqrt{7}, 0)$.

The foci are at $ \mathbf{(\sqrt{7}, 0)} $ and $ \mathbf{(-\sqrt{7}, 0)} $.

Coordinates of the vertices: $(\pm a, 0) = (\pm 4, 0)$.

The vertices are at $ \mathbf{(4, 0)} $ and $ \mathbf{(-4, 0)} $.

Length of major axis: $2a = 2 \times 4 = \mathbf{8}$ units.

Length of minor axis: $2b = 2 \times 3 = \mathbf{6}$ units.

Eccentricity: $e = \frac{c}{a} = \frac{\sqrt{7}}{4} = \mathbf{\frac{\sqrt{7}}{4}}$.

Length of latus rectum: $ \frac{2b^2}{a} = \frac{2 \times 9}{4} = \frac{18}{4} = \frac{9}{2} = \mathbf{4.5}$ units.

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} $ or $ 4.5 $

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 major axis, 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 of the standard form $ \frac{x^2}{b^2} + \frac{y^2}{a^2} = 1 $ (since the denominator of the $y^2$ term is larger, indicating the major axis is along the y-axis).

Comparing equation (1) with the standard form, we have:

From (2) and (3), we get $a = \sqrt{100} = 10$ and $b = \sqrt{25} = 5$.

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

The coordinates of the foci are $ (0, \pm c) $ and the coordinates of the vertices are $ (0, \pm a) $, where $c^2 = a^2 - b^2$.

Let's find the value of $c$:

Substitute the values of $a^2$ and $b^2$ from (2) and (3) into (4):

$c^2 = 100 - 25$

$c^2 = 75$

$c = \sqrt{75} = \sqrt{25 \times 3} = 5\sqrt{3} $

Now we can find the required properties:

Coordinates of the foci: $ (0, \pm c) = \mathbf{(0, \pm 5\sqrt{3})} $.

The foci are at $ (0, 5\sqrt{3}) $ and $ (0, -5\sqrt{3}) $.

Coordinates of the vertices: $ (0, \pm a) = \mathbf{(0, \pm 10)} $.

The vertices are at $ (0, 10) $ and $ (0, -10) $.

Length of major axis: $2a = 2 \times 10 = \mathbf{20}$ units.

Length of minor axis: $2b = 2 \times 5 = \mathbf{10}$ units.

Eccentricity: $e = \frac{c}{a} = \frac{5\sqrt{3}}{10} = \mathbf{\frac{\sqrt{3}}{2}}$.

Length of latus rectum: $ \frac{2b^2}{a} = \frac{2 \times 25}{10} = \frac{50}{10} = \mathbf{5}$ units.

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 major axis, 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 of the standard form $ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $.

Comparing equation (1) with the standard form, we have:

From (2) and (3), we get $a = \sqrt{49} = 7$ and $b = \sqrt{36} = 6$.

Since $a^2 = 49 > b^2 = 36$, the major axis of the ellipse is along the x-axis.

The coordinates of the foci are $ (\pm c, 0) $ and the coordinates of the vertices are $ (\pm a, 0) $, where $c^2 = a^2 - b^2$.

Let's find the value of $c$:

Substitute the values of $a^2$ and $b^2$ from (2) and (3) into (4):

$c^2 = 49 - 36$

$c^2 = 13$

$c = \sqrt{13} $

Now we can find the required properties:

Coordinates of the foci: $(\pm c, 0) = (\pm \sqrt{13}, 0)$.

The foci are at $ \mathbf{(\sqrt{13}, 0)} $ and $ \mathbf{(-\sqrt{13}, 0)} $.

Coordinates of the vertices: $(\pm a, 0) = (\pm 7, 0)$.

The vertices are at $ \mathbf{(7, 0)} $ and $ \mathbf{(-7, 0)} $.

Length of major axis: $2a = 2 \times 7 = \mathbf{14}$ units.

Length of minor axis: $2b = 2 \times 6 = \mathbf{12}$ units.

Eccentricity: $e = \frac{c}{a} = \frac{\sqrt{13}}{7} = \mathbf{\frac{\sqrt{13}}{7}}$.

Length of latus rectum: $ \frac{2b^2}{a} = \frac{2 \times 36}{7} = \frac{72}{7}$ units.

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 major axis, 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 of the standard form $ \frac{x^2}{b^2} + \frac{y^2}{a^2} = 1 $ (since the denominator of the $y^2$ term is larger, indicating the major axis is along the y-axis).

Comparing equation (1) with the standard form, we have:

From (2) and (3), we get $a = \sqrt{400} = 20$ and $b = \sqrt{100} = 10$.

Since $a^2 = 400 > b^2 = 100$, the major axis of the ellipse is along the y-axis.

The coordinates of the foci are $ (0, \pm c) $ and the coordinates of the vertices are $ (0, \pm a) $, where $c^2 = a^2 - b^2$.

Let's find the value of $c$:

Substitute the values of $a^2$ and $b^2$ from (2) and (3) into (4):

$c^2 = 400 - 100$

$c^2 = 300$

$c = \sqrt{300} = \sqrt{100 \times 3} = 10\sqrt{3} $

Now we can find the required properties:

Coordinates of the foci: $ (0, \pm c) = \mathbf{(0, \pm 10\sqrt{3})} $.

The foci are at $ (0, 10\sqrt{3}) $ and $ (0, -10\sqrt{3}) $.

Coordinates of the vertices: $ (0, \pm a) = \mathbf{(0, \pm 20)} $.

The vertices are at $ (0, 20) $ and $ (0, -20) $.

Length of major axis: $2a = 2 \times 20 = \mathbf{40}$ units.

Length of minor axis: $2b = 2 \times 10 = \mathbf{20}$ units.

Eccentricity: $e = \frac{c}{a} = \frac{10\sqrt{3}}{20} = \mathbf{\frac{\sqrt{3}}{2}}$.

Length of latus rectum: $ \frac{2b^2}{a} = \frac{2 \times 100}{20} = \frac{200}{20} = \mathbf{10}$ units.

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 major axis, 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 into the standard form $ \frac{x^2}{B^2} + \frac{y^2}{A^2} = 1 $ or $ \frac{x^2}{A^2} + \frac{y^2}{B^2} = 1 $, we need to divide both sides of the equation by 144.

$ \frac{36x^2}{144} + \frac{4y^2}{144} = \frac{144}{144} $

Simplify the fractions:

$ \frac{\cancel{36}^{1}x^2}{\cancel{144}_{4}} + \frac{\cancel{4}^{1}y^2}{\cancel{144}_{36}} = 1 $

This gives the standard form of the ellipse equation:

This equation is of the standard form $ \frac{x^2}{b^2} + \frac{y^2}{a^2} = 1 $ (since the denominator of the $y^2$ term is larger, indicating the major axis is along the y-axis).

Comparing equation (1) with the standard form, we have:

From (2) and (3), we get $a = \sqrt{36} = 6$ and $b = \sqrt{4} = 2$.

Since $a^2 = 36 > b^2 = 4$, the major axis of the ellipse is along the y-axis.

The coordinates of the foci are $ (0, \pm c) $ and the coordinates of the vertices are $ (0, \pm a) $, where $c^2 = a^2 - b^2$.

Let's find the value of $c$:

Substitute the values of $a^2$ and $b^2$ from (2) and (3) into (4):

$c^2 = 36 - 4$

$c^2 = 32$

$c = \sqrt{32} = \sqrt{16 \times 2} = 4\sqrt{2} $

Now we can find the required properties:

Coordinates of the foci: $ (0, \pm c) = \mathbf{(0, \pm 4\sqrt{2})} $.

The foci are at $ (0, 4\sqrt{2}) $ and $ (0, -4\sqrt{2}) $.

Coordinates of the vertices: $ (0, \pm a) = \mathbf{(0, \pm 6)} $.

The vertices are at $ (0, 6) $ and $ (0, -6) $.

Length of major axis: $2a = 2 \times 6 = \mathbf{12}$ units.

Length of minor axis: $2b = 2 \times 2 = \mathbf{4}$ units.

Eccentricity: $e = \frac{c}{a} = \frac{4\sqrt{2}}{6} = \mathbf{\frac{2\sqrt{2}}{3}}$.

Length of latus rectum: $ \frac{2b^2}{a} = \frac{2 \times 4}{6} = \frac{8}{6} = \frac{4}{3}$ units.

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 major axis, 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 into the standard form $ \frac{x^2}{B^2} + \frac{y^2}{A^2} = 1 $ or $ \frac{x^2}{A^2} + \frac{y^2}{B^2} = 1 $, we need to divide both sides of the equation by 16.

$ \frac{16x^2}{16} + \frac{y^2}{16} = \frac{16}{16} $

Simplify the fractions:

$ \frac{\cancel{16}^{1}x^2}{\cancel{16}_{1}} + \frac{y^2}{16} = 1 $

This gives the standard form of the ellipse equation:

This equation is of the standard form $ \frac{x^2}{b^2} + \frac{y^2}{a^2} = 1 $ (since the denominator of the $y^2$ term is larger, indicating the major axis is along the y-axis).

Comparing equation (1) with the standard form, we have:

From (2) and (3), we get $a = \sqrt{16} = 4$ and $b = \sqrt{1} = 1$.

Since $a^2 = 16 > b^2 = 1$, the major axis of the ellipse is along the y-axis.

The coordinates of the foci are $ (0, \pm c) $ and the coordinates of the vertices are $ (0, \pm a) $, where $c^2 = a^2 - b^2$.

Let's find the value of $c$:

Substitute the values of $a^2$ and $b^2$ from (2) and (3) into (4):

$c^2 = 16 - 1$

$c^2 = 15$

$c = \sqrt{15} $

Now we can find the required properties:

Coordinates of the foci: $ (0, \pm c) = \mathbf{(0, \pm \sqrt{15})} $.

The foci are at $ (0, \sqrt{15}) $ and $ (0, -\sqrt{15}) $.

Coordinates of the vertices: $ (0, \pm a) = \mathbf{(0, \pm 4)} $.

The vertices are at $ (0, 4) $ and $ (0, -4) $.

Length of major axis: $2a = 2 \times 4 = \mathbf{8}$ units.

Length of minor axis: $2b = 2 \times 1 = \mathbf{2}$ units.

Eccentricity: $e = \frac{c}{a} = \mathbf{\frac{\sqrt{15}}{4}}$.

Length of latus rectum: $ \frac{2b^2}{a} = \frac{2 \times 1^2}{4} = \frac{2}{4} = \frac{1}{2} = \mathbf{0.5}$ units.

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} $ or $ 0.5 $

Given:

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

To Find:

The coordinates of the foci, the vertices, the length of major axis, 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 into the standard form $ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $ or $ \frac{x^2}{b^2} + \frac{y^2}{a^2} = 1 $, we need to divide both sides of the equation by 36.

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

Simplify the fractions:

$ \frac{\cancel{4}^{1}x^2}{\cancel{36}_{9}} + \frac{\cancel{9}^{1}y^2}{\cancel{36}_{4}} = 1 $

This gives the standard form of the ellipse equation:

This equation is of the standard form $ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $.

Comparing equation (1) with the standard form, we have:

From (2) and (3), we get $a = \sqrt{9} = 3$ and $b = \sqrt{4} = 2$.

Since $a^2 = 9 > b^2 = 4$, the major axis of the ellipse is along the x-axis.

The coordinates of the foci are $ (\pm c, 0) $ and the coordinates of the vertices are $ (\pm a, 0) $, where $c^2 = a^2 - b^2$.

Let's find the value of $c$:

Substitute the values of $a^2$ and $b^2$ from (2) and (3) into (4):

$c^2 = 9 - 4$

$c^2 = 5$

$c = \sqrt{5} $

Now we can find the required properties:

Coordinates of the foci: $(\pm c, 0) = (\pm \sqrt{5}, 0)$.

The foci are at $ \mathbf{(\sqrt{5}, 0)} $ and $ \mathbf{(-\sqrt{5}, 0)} $.

Coordinates of the vertices: $(\pm a, 0) = (\pm 3, 0)$.

The vertices are at $ \mathbf{(3, 0)} $ and $ \mathbf{(-3, 0)} $.

Length of major axis: $2a = 2 \times 3 = \mathbf{6}$ units.

Length of minor axis: $2b = 2 \times 2 = \mathbf{4}$ units.

Eccentricity: $e = \frac{c}{a} = \frac{\sqrt{5}}{3} = \mathbf{\frac{\sqrt{5}}{3}}$.

Length of latus rectum: $ \frac{2b^2}{a} = \frac{2 \times 4}{3} = \frac{8}{3}$ units.

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:

Since the vertices are $(\pm 5, 0)$ and the foci are $(\pm 4, 0)$, the center of the ellipse is at the origin $ (0, 0) $.

Also, the vertices and foci lie on the x-axis. This indicates that the major axis of the ellipse is along the x-axis.

The standard form of the equation of an ellipse with the center at the origin and major axis along the x-axis is:

where $a$ is the length of the semi-major axis and $b$ is the length of the semi-minor axis, and $a > b$.

The coordinates of the vertices of such an ellipse are $(\pm a, 0)$.

Comparing the given vertices $(\pm 5, 0)$ with $(\pm a, 0)$, we get:

So, $a^2 = 5^2 = 25$.

The coordinates of the foci of such an ellipse are $(\pm c, 0)$, where $c$ is the distance from the center to each focus.

Comparing the given foci $(\pm 4, 0)$ with $(\pm c, 0)$, we get:

So, $c^2 = 4^2 = 16$.

We know the relationship between $a$, $b$, and $c$ for an ellipse is $c^2 = a^2 - b^2$.

We can use this relation to find $b^2$:

Substitute the values of $a^2 = 25$ (from 2) and $c^2 = 16$ (from 3) into (4):

$b^2 = 25 - 16$

$b^2 = 9$

Now, substitute the values of $a^2 = 25$ and $b^2 = 9$ into the standard equation of the ellipse (1):

Thus, the equation of the ellipse is $ \mathbf{\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:

Since the vertices are $(0, \pm 13)$ and the foci are $(0, \pm 5)$, the center of the ellipse is at the origin $ (0, 0) $.

Also, the vertices and foci lie on the y-axis. This indicates that the major axis of the ellipse is along the y-axis.

The standard form of the equation of an ellipse with the center at the origin and major axis along the y-axis is:

where $a$ is the length of the semi-major axis and $b$ is the length of the semi-minor axis, and $a > b$.

The coordinates of the vertices of such an ellipse are $(0, \pm a)$.

Comparing the given vertices $(0, \pm 13)$ with $(0, \pm a)$, we get:

So, $a^2 = 13^2 = 169$.

The coordinates of the foci of such an ellipse are $(0, \pm c)$, where $c$ is the distance from the center to each focus.

Comparing the given foci $(0, \pm 5)$ with $(0, \pm c)$, we get:

So, $c^2 = 5^2 = 25$.

We know the relationship between $a$, $b$, and $c$ for an ellipse is $c^2 = a^2 - b^2$.

We can use this relation to find $b^2$:

Substitute the values of $a^2 = 169$ (from 2) and $c^2 = 25$ (from 3) into (4):

$b^2 = 169 - 25$

$b^2 = 144$

Now, substitute the values of $a^2 = 169$ and $b^2 = 144$ into the standard equation of the ellipse (1):

Thus, the equation of the ellipse is $ \mathbf{\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:

Since the vertices are $(\pm 6, 0)$ and the foci are $(\pm 4, 0)$, the center of the ellipse is at the origin $ (0, 0) $.

Also, the vertices and foci lie on the x-axis. This indicates that the major axis of the ellipse is along the x-axis.

The standard form of the equation of an ellipse with the center at the origin and major axis along the x-axis is:

where $a$ is the length of the semi-major axis and $b$ is the length of the semi-minor axis, and $a > b$.

The coordinates of the vertices of such an ellipse are $(\pm a, 0)$.

Comparing the given vertices $(\pm 6, 0)$ with $(\pm a, 0)$, we get:

So, $a^2 = 6^2 = 36$.

The coordinates of the foci of such an ellipse are $(\pm c, 0)$, where $c$ is the distance from the center to each focus.

Comparing the given foci $(\pm 4, 0)$ with $(\pm c, 0)$, we get:

So, $c^2 = 4^2 = 16$.

We know the relationship between $a$, $b$, and $c$ for an ellipse is $c^2 = a^2 - b^2$.

We can use this relation to find $b^2$:

Substitute the values of $a^2 = 36$ (from 2) and $c^2 = 16$ (from 3) into (4):

$b^2 = 36 - 16$

$b^2 = 20$

Now, substitute the values of $a^2 = 36$ and $b^2 = 20$ into the standard equation of the ellipse (1):

Thus, the equation of the ellipse is $ \mathbf{\frac{x^2}{36} + \frac{y^2}{20} = 1} $.

Given:

The ends of the major axis of the ellipse are $(\pm 3, 0)$.

The ends of the minor axis of the ellipse are $(0, \pm 2)$.

To Find:

The equation of the ellipse.

Solution:

The ends of the major axis are given as $(\pm 3, 0)$. Since these points lie on the x-axis, the major axis is along the x-axis.

The center of the ellipse is the midpoint of the major axis, which is $ (\frac{3+(-3)}{2}, \frac{0+0}{2}) = (0, 0) $. So the center is at the origin.

The standard form of the equation of an ellipse with the center at the origin and major axis along the x-axis is:

where $a$ is the length of the semi-major axis and $b$ is the length of the semi-minor axis, with $a > b$.

The ends of the major axis are $(\pm a, 0)$. Comparing the given ends $(\pm 3, 0)$ with $(\pm a, 0)$, we find the length of the semi-major axis:

So, $a^2 = 3^2 = 9$.

The ends of the minor axis are $(0, \pm b)$. Comparing the given ends $(0, \pm 2)$ with $(0, \pm b)$, we find the length of the semi-minor axis:

So, $b^2 = 2^2 = 4$.

Substitute the values of $a^2 = 9$ (from 2) and $b^2 = 4$ (from 3) into the standard equation of the ellipse (1):

Thus, the equation of the ellipse is $ \mathbf{\frac{x^2}{9} + \frac{y^2}{4} = 1} $.

Given:

The ends of the major axis of the ellipse are $(0, \pm \sqrt{5})$.

The ends of the minor axis of the ellipse are $(\pm 1, 0)$.

To Find:

The equation of the ellipse.

Solution:

The ends of the major axis are given as $(0, \pm \sqrt{5})$. Since these points lie on the y-axis, the major axis is along the y-axis.

The center of the ellipse is the midpoint of the major axis, which is $ (\frac{0+0}{2}, \frac{\sqrt{5}+(-\sqrt{5})}{2}) = (0, 0) $. So the center is at the origin.

The standard form of the equation of an ellipse with the center at the origin and major axis along the y-axis is:

where $a$ is the length of the semi-major axis and $b$ is the length of the semi-minor axis, with $a > b$.

The ends of the major axis are $(0, \pm a)$. Comparing the given ends $(0, \pm \sqrt{5})$ with $(0, \pm a)$, we find the length of the semi-major axis:

So, $a^2 = (\sqrt{5})^2 = 5$.

The ends of the minor axis are $(\pm b, 0)$. Comparing the given ends $(\pm 1, 0)$ with $(\pm b, 0)$, we find the length of the semi-minor axis:

So, $b^2 = 1^2 = 1$.

Substitute the values of $a^2 = 5$ (from 2) and $b^2 = 1$ (from 3) into the standard equation of the ellipse (1):

This can also be written as $ x^2 + \frac{y^2}{5} = 1 $.

Thus, the equation of the ellipse is $ \mathbf{x^2 + \frac{y^2}{5} = 1} $ or $ \mathbf{\frac{x^2}{1} + \frac{y^2}{5} = 1} $.

Given:

Length of the major axis = 26.

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

To Find:

The equation of the ellipse.

Solution:

Since the foci are located at $(\pm 5, 0)$, they lie on the x-axis.

This means the center of the ellipse is at the origin $(0, 0)$ and the major axis is along the x-axis.

The standard form of the equation of an ellipse with the center at the origin and major axis along the x-axis is:

where $a$ is the length of the semi-major axis and $b$ is the length of the semi-minor axis, and $a > b$.

The length of the major axis is given as 26. For an ellipse with the major axis along the x-axis, the length of the major axis is $2a$.

So, we have:

Dividing by 2, we get:

So, $a^2 = 13^2 = 169$.

The coordinates of the foci of such an ellipse are $(\pm c, 0)$, where $c$ is the distance from the center to each focus.

Comparing the given foci $(\pm 5, 0)$ with $(\pm c, 0)$, we get:

So, $c^2 = 5^2 = 25$.

We use the relationship between $a$, $b$, and $c$ for an ellipse, which is $c^2 = a^2 - b^2$.

We can rearrange this to find $b^2$:

Substitute the values of $a^2 = 169$ (from 2) and $c^2 = 25$ (from 3) into (4):

$b^2 = 169 - 25$

$b^2 = 144$

Now, substitute the values of $a^2 = 169$ and $b^2 = 144$ into the standard equation of the ellipse (1):

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

Given:

Length of the minor axis = 16.

The coordinates of the foci are $(0, \pm 6)$.

To Find:

The equation of the ellipse.

Solution:

Since the foci are located at $(0, \pm 6)$, they lie on the y-axis.

This means the center of the ellipse is at the origin $(0, 0)$ and the major axis is along the y-axis.

The standard form of the equation of an ellipse with the center at the origin and major axis along the y-axis is:

where $a$ is the length of the semi-major axis and $b$ is the length of the semi-minor axis, and $a > b$.

The coordinates of the foci of such an ellipse are $(0, \pm c)$, where $c$ is the distance from the center to each focus.

Comparing the given foci $(0, \pm 6)$ with $(0, \pm c)$, we get:

So, $c^2 = 6^2 = 36$.

The length of the minor axis is given as 16. For an ellipse with the major axis along the y-axis, the length of the minor axis is $2b$.

So, we have:

Dividing by 2, we get:

So, $b^2 = 8^2 = 64$.

We use the relationship between $a$, $b$, and $c$ for an ellipse, which is $c^2 = a^2 - b^2$.

We can rearrange this to find $a^2$:

Substitute the values of $b^2 = 64$ (from 3) and $c^2 = 36$ (from 2) into (4):

$a^2 = 64 + 36$

$a^2 = 100$

So, $a = \sqrt{100} = 10$. (Note that $a=10 > b=8$, confirming the major axis is along the y-axis).

Now, substitute the values of $a^2 = 100$ and $b^2 = 64$ into the standard equation of the ellipse (1):

Thus, the equation of the ellipse is $ \mathbf{\frac{x^2}{64} + \frac{y^2}{100} = 1} $.

Given:

The coordinates of the foci are $(\pm 3, 0)$.

The length of the semi-major axis $a = 4$.

To Find:

The equation of the ellipse.

Solution:

Since the foci are located at $(\pm 3, 0)$, they lie on the x-axis.

This means the center of the ellipse is at the origin $(0, 0)$ and the major axis is along the x-axis.

The standard form of the equation of an ellipse with the center at the origin and major axis along the x-axis is:

where $a$ is the length of the semi-major axis and $b$ is the length of the semi-minor axis, and $a > b$.

We are given the length of the semi-major axis:

So, $a^2 = 4^2 = 16$.

The coordinates of the foci of such an ellipse are $(\pm c, 0)$, where $c$ is the distance from the center to each focus.

Comparing the given foci $(\pm 3, 0)$ with $(\pm c, 0)$, we get:

So, $c^2 = 3^2 = 9$.

We use the relationship between $a$, $b$, and $c$ for an ellipse, which is $c^2 = a^2 - b^2$.

We can rearrange this to find $b^2$:

Substitute the values of $a^2 = 16$ (from 2) and $c^2 = 9$ (from 3) into (4):

$b^2 = 16 - 9$

$b^2 = 7$

Now, substitute the values of $a^2 = 16$ and $b^2 = 7$ into the standard equation of the ellipse (1):

Thus, the equation of the ellipse is $ \mathbf{\frac{x^2}{16} + \frac{y^2}{7} = 1} $.

Given:

The length of the semi-minor axis $ b = 3 $.

The distance from the center to the foci $ c = 4 $.

The center of the ellipse is at the origin $ (0, 0) $.

The foci are on the x-axis.

To Find:

The equation of the ellipse.

Solution:

Since the foci are on the x-axis and the center is at the origin, the major axis of the ellipse is along the x-axis.

The standard form of the equation of an ellipse with the center at the origin and major axis along the x-axis is:

where $a$ is the length of the semi-major axis and $b$ is the length of the semi-minor axis, and $a > b$.

We are given the length of the semi-minor axis:

So, $b^2 = 3^2 = 9$.

We are given the distance from the center to the foci:

So, $c^2 = 4^2 = 16$.

We use the relationship between $a$, $b$, and $c$ for an ellipse, which is $c^2 = a^2 - b^2$.

We can rearrange this to find $a^2$:

Substitute the values of $b^2 = 9$ (from 2) and $c^2 = 16$ (from 3) into (4):

$a^2 = 9 + 16$

$a^2 = 25$

So, $a = \sqrt{25} = 5$. (Note that $a=5 > b=3$, confirming the major axis is along the x-axis).

Now, substitute the values of $a^2 = 25$ and $b^2 = 9$ into the standard equation of the ellipse (1):

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

Given:

The center of the ellipse is at the origin $(0, 0)$.

The major axis 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 center of the ellipse is at the origin and the major axis is on the y-axis, the standard form of the equation of the ellipse is:

where $a$ is the length of the semi-major axis and $b$ is the length of the semi-minor axis, and $a > b$.

The ellipse passes through the point $ (3, 2) $. Substituting $x=3$ and $y=2$ into equation (1), we get:

This simplifies to:

The ellipse also passes through the point $ (1, 6) $. Substituting $x=1$ and $y=6$ into equation (1), we get:

This simplifies to:

We now have a system of two linear equations in terms of $ \frac{1}{b^2} $ and $ \frac{1}{a^2} $.

Let $ X = \frac{1}{b^2} $ and $ Y = \frac{1}{a^2} $.

The system of equations (3) and (5) becomes:

$9X + 4Y = 1$

$X + 36Y = 1$

From the second equation, we can express $X$ in terms of $Y$:

Substitute equation (6) into the first equation:

$9(1 - 36Y) + 4Y = 1$

$9 - 324Y + 4Y = 1$

$9 - 320Y = 1$

$9 - 1 = 320Y$

$8 = 320Y$

$Y = \frac{8}{320} $

So, $ \frac{1}{a^2} = \frac{1}{40} $, which means $ a^2 = 40 $.

Now substitute the value of $Y$ from (7) into equation (6) to find $X$:

$X = 1 - 36 \left(\frac{1}{40}\right) $

$X = 1 - \frac{36}{40} $

$X = 1 - \frac{9}{10} $

$X = \frac{10 - 9}{10} $

So, $ \frac{1}{b^2} = \frac{1}{10} $, which means $ b^2 = 10 $.

We have $a^2 = 40$ and $b^2 = 10$. Since $40 > 10$, $a^2 > b^2$, which confirms that the major axis is along the y-axis.

Substitute the values of $ a^2 = 40 $ and $ b^2 = 10 $ into the standard equation of the ellipse (1):

Thus, the equation of the ellipse is $ \mathbf{\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 of the ellipse is on the x-axis and the center is implicitly assumed to be at the origin, the standard form of the equation of the ellipse is:

where $a$ is the length of the semi-major axis and $b$ is the length of the semi-minor axis, and $a > b$.

The ellipse passes through the point $ (4, 3) $. Substituting $x=4$ and $y=3$ into equation (1), we get:

This simplifies to:

The ellipse also passes through the point $ (6, 2) $. Substituting $x=6$ and $y=2$ into equation (1), we get:

This simplifies to:

We now have a system of two linear equations in terms of $ \frac{1}{a^2} $ and $ \frac{1}{b^2} $.

Let $ X = \frac{1}{a^2} $ and $ Y = \frac{1}{b^2} $.

The system of equations (3) and (5) becomes:

To solve for $X$ and $Y$, we can multiply equation (6) by 4 and equation (7) by 9:

$4 \times (16X + 9Y) = 4 \times 1 \implies 64X + 36Y = 4$

$9 \times (36X + 4Y) = 9 \times 1 \implies 324X + 36Y = 9$

Subtract the first resulting equation from the second:

$ (324X + 36Y) - (64X + 36Y) = 9 - 4 $

$ 324X - 64X = 5 $

$ 260X = 5 $

So, $ \frac{1}{a^2} = \frac{1}{52} $, which means $ a^2 = 52 $.

Substitute the value of $X$ from (8) into equation (6):

$16\left(\frac{1}{52}\right) + 9Y = 1$

$ \frac{16}{52} + 9Y = 1 $

$ \frac{4}{13} + 9Y = 1 $

$ 9Y = 1 - \frac{4}{13} $

$ 9Y = \frac{13 - 4}{13} = \frac{9}{13} $

So, $ \frac{1}{b^2} = \frac{1}{13} $, which means $ b^2 = 13 $.

We have $a^2 = 52$ and $b^2 = 13$. Since $52 > 13$, $a^2 > b^2$, which confirms that the major axis is along the x-axis.

Substitute the values of $ a^2 = 52 $ and $ b^2 = 13 $ into the standard equation of the ellipse (1):

Thus, the equation of the ellipse is $ \mathbf{\frac{x^2}{52} + \frac{y^2}{13} = 1} $.

(i)

Given:

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

To Find:

The coordinates of the foci and 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 of the standard form $ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 $.

Comparing equation (1) with the standard form, we have:

From (2) and (3), we get $a = \sqrt{9} = 3$ and $b = \sqrt{16} = 4$.

Since the $x^2$ term is positive, the transverse axis is along the x-axis.

The coordinates of the foci are $ (\pm c, 0) $ and the coordinates of the vertices are $ (\pm a, 0) $, where $c^2 = a^2 + b^2$.

Let's find the value of $c$:

Substitute the values of $a^2$ and $b^2$ from (2) and (3) into (4):

$c^2 = 9 + 16$

$c^2 = 25$

$c = \sqrt{25} = 5 $

Now we can find the required properties:

Coordinates of the foci: $(\pm c, 0) = (\pm 5, 0)$.

The foci are at $ \mathbf{(5, 0)} $ and $ \mathbf{(-5, 0)} $.

Coordinates of the vertices: $(\pm a, 0) = (\pm 3, 0)$.

The vertices are at $ \mathbf{(3, 0)} $ and $ \mathbf{(-3, 0)} $.

Eccentricity: $e = \frac{c}{a} = \frac{5}{3} = \mathbf{\frac{5}{3}}$.

Length of latus rectum: $ \frac{2b^2}{a} = \frac{2 \times 16}{3} = \frac{32}{3}$ units.

Summary of Results for (i):

Foci: $ (\pm 5, 0) $

Vertices: $ (\pm 3, 0) $

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

Length of Latus Rectum: $ \frac{32}{3} $

(ii)

Given:

The equation of the hyperbola is $ y^2 – 16x^2 = 16 $.

To Find:

The coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the given hyperbola.

Solution:

The given equation of the hyperbola is $ y^2 – 16x^2 = 16 $.

To convert this into the standard form $ \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 $, we need to divide both sides of the equation by 16.

$ \frac{y^2}{16} - \frac{16x^2}{16} = \frac{16}{16} $

Simplify the fractions:

$ \frac{y^2}{16} - \frac{\cancel{16}^{1}x^2}{\cancel{16}_{1}} = 1 $

This gives the standard form of the hyperbola equation:

This equation is of the standard form $ \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 $ (since the $y^2$ term is positive, indicating the transverse axis is along the y-axis).

Comparing equation (5) with the standard form, we have:

From (6) and (7), we get $a = \sqrt{16} = 4$ and $b = \sqrt{1} = 1$.

Since the transverse axis is along the y-axis, the coordinates of the foci are $ (0, \pm c) $ and the coordinates of the vertices are $ (0, \pm a) $, where $c^2 = a^2 + b^2$.

Let's find the value of $c$:

Substitute the values of $a^2$ and $b^2$ from (6) and (7) into (8):

$c^2 = 16 + 1$

$c^2 = 17$

$c = \sqrt{17} $

Now we can find the required properties:

Coordinates of the foci: $ (0, \pm c) = \mathbf{(0, \pm \sqrt{17})} $.

The foci are at $ (0, \sqrt{17}) $ and $ (0, -\sqrt{17}) $.

Coordinates of the vertices: $ (0, \pm a) = \mathbf{(0, \pm 4)} $.

The vertices are at $ (0, 4) $ and $ (0, -4) $.

Eccentricity: $e = \frac{c}{a} = \frac{\sqrt{17}}{4} = \mathbf{\frac{\sqrt{17}}{4}}$.

Length of latus rectum: $ \frac{2b^2}{a} = \frac{2 \times 1^2}{4} = \frac{2}{4} = \frac{1}{2} = \mathbf{0.5}$ units.

Summary of Results for (ii):

Foci: $ (0, \pm \sqrt{17}) $

Vertices: $ (0, \pm 4) $

Eccentricity: $ \frac{\sqrt{17}}{4} $

Length of Latus Rectum: $ \frac{1}{2} $ or $ 0.5 $

Given:

The coordinates of the foci are $(0, \pm 3)$.

The coordinates of the vertices are $ \left( 0, \pm \frac{\sqrt{11}}{2} \right) $.

To Find:

The equation of the hyperbola.

Solution:

Since the foci are located at $(0, \pm 3)$ and the vertices are at $ \left( 0, \pm \frac{\sqrt{11}}{2} \right) $, they both lie on the y-axis.

This means the center of the hyperbola is at the origin $(0, 0)$ and the transverse axis is along the y-axis.

The standard form of the equation of a hyperbola with the center at the origin and transverse axis along the y-axis is:

where $a$ is the distance from the center to each vertex along the transverse axis, and $b$ is the distance from the center to each end of the conjugate axis.

The coordinates of the vertices of such a hyperbola are $(0, \pm a)$.

Comparing the given vertices $ \left( 0, \pm \frac{\sqrt{11}}{2} \right) $ with $(0, \pm a)$, we get:

So, $a^2 = \left(\frac{\sqrt{11}}{2}\right)^2 = \frac{11}{4}$.

The coordinates of the foci of such a hyperbola are $(0, \pm c)$, where $c$ is the distance from the center to each focus.

Comparing the given foci $(0, \pm 3)$ with $(0, \pm c)$, we get:

So, $c^2 = 3^2 = 9$.

We use the relationship between $a$, $b$, and $c$ for a hyperbola, which is $c^2 = a^2 + b^2$.

We can rearrange this to find $b^2$:

Substitute the values of $c^2 = 9$ (from 3) and $a^2 = \frac{11}{4}$ (from 2) into (4):

$b^2 = 9 - \frac{11}{4} $

$b^2 = \frac{9 \times 4}{4} - \frac{11}{4} $

$b^2 = \frac{36}{4} - \frac{11}{4} $

$b^2 = \frac{36 - 11}{4} $

$b^2 = \frac{25}{4} $

Now, substitute the values of $a^2 = \frac{11}{4}$ and $b^2 = \frac{25}{4}$ into the standard equation of the hyperbola (1):

$ \frac{y^2}{\frac{11}{4}} - \frac{x^2}{\frac{25}{4}} = 1 $

Simplify the fractions by multiplying the numerator and denominator of each term by 4:

Thus, the equation of the hyperbola is $ \mathbf{\frac{4y^2}{11} - \frac{4x^2}{25} = 1} $.

Given:

The coordinates of the foci are $(0, \pm 12)$.

The length of the latus rectum is 36.

To Find:

The equation of the hyperbola.

Solution:

Since the foci are located at $(0, \pm 12)$, they lie on the y-axis.

This means the center of the hyperbola is at the origin $(0, 0)$ and the transverse axis is along the y-axis.

The standard form of the equation of a hyperbola with the center at the origin and transverse axis along the y-axis is:

where $a$ is the distance from the center to each vertex along the transverse axis, and $b$ is the distance from the center to each end of the conjugate axis.

The coordinates of the foci of such a hyperbola are $(0, \pm c)$, where $c$ is the distance from the center to each focus.

Comparing the given foci $(0, \pm 12)$ with $(0, \pm c)$, we get:

So, $c^2 = 12^2 = 144$.

The length of the latus rectum of a hyperbola with the transverse axis along the y-axis is given by the formula $ \frac{2b^2}{a} $. We are given that this length is 36.

From equation (3), we can express $2b^2$ in terms of $a$:

$2b^2 = 36a$

We use the relationship between $a$, $b$, and $c$ for a hyperbola, which is $c^2 = a^2 + b^2$.

Substitute the value of $c^2 = 144$ (from 2) and $b^2 = 18a$ (from 4) into (5):

$144 = a^2 + 18a$

Rearrange the equation into a quadratic equation in $a$:

$a^2 + 18a - 144 = 0$

We can solve this quadratic equation for $a$ by factoring. We look for two numbers that multiply to -144 and add up to 18. These numbers are 24 and -6.

$a^2 + 24a - 6a - 144 = 0$

$a(a + 24) - 6(a + 24) = 0$

$(a - 6)(a + 24) = 0$

This gives two possible values for $a$: $a = 6$ or $a = -24$.

Since $a$ represents a distance, it must be positive. Therefore, we take $a = 6$.

So, $a^2 = 6^2 = 36$.

Now, substitute the value of $a = 6$ into equation (4) to find $b^2$:

$b^2 = 18 \times 6$

Substitute the values of $a^2 = 36$ (from 6) and $b^2 = 108$ (from 7) into the standard equation of the hyperbola (1):

Thus, the equation of the hyperbola is $ \mathbf{\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 and 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 of the standard form $ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 $.

Comparing equation (1) with the standard form, we have:

From (2) and (3), we get $a = \sqrt{16} = 4$ and $b = \sqrt{9} = 3$.

Since the $x^2$ term is positive, the transverse axis is along the x-axis.

The coordinates of the foci are $ (\pm c, 0) $ and the coordinates of the vertices are $ (\pm a, 0) $, where $c^2 = a^2 + b^2$.

Let's find the value of $c$:

Substitute the values of $a^2$ and $b^2$ from (2) and (3) into (4):

$c^2 = 16 + 9$

$c^2 = 25$

$c = \sqrt{25} = 5 $

Now we can find the required properties:

Coordinates of the foci: $(\pm c, 0) = (\pm 5, 0)$.

The foci are at $ \mathbf{(5, 0)} $ and $ \mathbf{(-5, 0)} $.

Coordinates of the vertices: $(\pm a, 0) = (\pm 4, 0)$.

The vertices are at $ \mathbf{(4, 0)} $ and $ \mathbf{(-4, 0)} $.

Eccentricity: $e = \frac{c}{a} = \frac{5}{4} = \mathbf{\frac{5}{4}}$.

Length of latus rectum: $ \frac{2b^2}{a} = \frac{2 \times 9}{4} = \frac{18}{4} = \frac{9}{2} = \mathbf{4.5}$ units.

Summary of Results:

Foci: $ (\pm 5, 0) $

Vertices: $ (\pm 4, 0) $

Eccentricity: $ \frac{5}{4} $

Length of Latus Rectum: $ \frac{9}{2} $ or $ 4.5 $

Given:

The equation of the hyperbola is $ \frac{y^{2}}{9} - \frac{x^{2}}{27} = 1 $.

To Find:

The coordinates of the foci and 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 of the standard form $ \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 $ (since the $y^2$ term is positive, indicating the transverse axis is along the y-axis).

Comparing equation (1) with the standard form, we have:

From (2) and (3), we get $a = \sqrt{9} = 3$ and $b = \sqrt{27} = \sqrt{9 \times 3} = 3\sqrt{3}$.

Since the transverse axis is along the y-axis, the coordinates of the foci are $ (0, \pm c) $ and the coordinates of the vertices are $ (0, \pm a) $, where $c^2 = a^2 + b^2$.

Let's find the value of $c$:

Substitute the values of $a^2$ and $b^2$ from (2) and (3) into (4):

$c^2 = 9 + 27$

$c^2 = 36$

$c = \sqrt{36} = 6 $

Now we can find the required properties:

Coordinates of the foci: $ (0, \pm c) = \mathbf{(0, \pm 6)} $.

The foci are at $ (0, 6) $ and $ (0, -6) $.

Coordinates of the vertices: $ (0, \pm a) = \mathbf{(0, \pm 3)} $.

The vertices are at $ (0, 3) $ and $ (0, -3) $.

Eccentricity: $e = \frac{c}{a} = \frac{6}{3} = \mathbf{2}$.

Length of latus rectum: $ \frac{2b^2}{a} = \frac{2 \times 27}{3} = \frac{54}{3} = \mathbf{18}$ units.

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 and 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 into the standard form $ \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 $, we need to divide both sides of the equation by 36.

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

Simplify the fractions:

$ \frac{\cancel{9}^{1}y^2}{\cancel{36}_{4}} - \frac{\cancel{4}^{1}x^2}{\cancel{36}_{9}} = 1 $

This gives the standard form of the hyperbola equation:

This equation is of the standard form $ \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 $ (since the $y^2$ term is positive, indicating the transverse axis is along the y-axis).

Comparing equation (1) with the standard form, we have:

From (2) and (3), we get $a = \sqrt{4} = 2$ and $b = \sqrt{9} = 3$.

Since the transverse axis is along the y-axis, the coordinates of the foci are $ (0, \pm c) $ and the coordinates of the vertices are $ (0, \pm a) $, where $c^2 = a^2 + b^2$.

Let's find the value of $c$:

Substitute the values of $a^2$ and $b^2$ from (2) and (3) into (4):

$c^2 = 4 + 9$

$c^2 = 13$

$c = \sqrt{13} $

Now we can find the required properties:

Coordinates of the foci: $ (0, \pm c) = \mathbf{(0, \pm \sqrt{13})} $.

The foci are at $ (0, \sqrt{13}) $ and $ (0, -\sqrt{13}) $.

Coordinates of the vertices: $ (0, \pm a) = \mathbf{(0, \pm 2)} $.

The vertices are at $ (0, 2) $ and $ (0, -2) $.

Eccentricity: $e = \frac{c}{a} = \frac{\sqrt{13}}{2} = \mathbf{\frac{\sqrt{13}}{2}}$.

Length of latus rectum: $ \frac{2b^2}{a} = \frac{2 \times 9}{2} = 9$ units.

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 and 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 into the standard form $ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 $, we need to divide both sides of the equation by 576.

$ \frac{16x^2}{576} - \frac{9y^2}{576} = \frac{576}{576} $

Simplify the fractions:

$ \frac{\cancel{16}^{1}x^2}{\cancel{576}_{36}} - \frac{\cancel{9}^{1}y^2}{\cancel{576}_{64}} = 1 $

(Note: $ 576 \div 16 = 36 $ and $ 576 \div 9 = 64 $)

This gives the standard form of the hyperbola equation:

This equation is of the standard form $ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 $ (since the $x^2$ term is positive, indicating the transverse axis is along the x-axis).

Comparing equation (1) with the standard form, we have:

From (2) and (3), we get $a = \sqrt{36} = 6$ and $b = \sqrt{64} = 8$.

Since the transverse axis is along the x-axis, the coordinates of the foci are $ (\pm c, 0) $ and the coordinates of the vertices are $ (\pm a, 0) $, where $c^2 = a^2 + b^2$.

Let's find the value of $c$:

Substitute the values of $a^2$ and $b^2$ from (2) and (3) into (4):

$c^2 = 36 + 64$

$c^2 = 100$

$c = \sqrt{100} = 10 $

Now we can find the required properties:

Coordinates of the foci: $(\pm c, 0) = (\pm 10, 0)$.

The foci are at $ \mathbf{(10, 0)} $ and $ \mathbf{(-10, 0)} $.

Coordinates of the vertices: $(\pm a, 0) = (\pm 6, 0)$.

The vertices are at $ \mathbf{(6, 0)} $ and $ \mathbf{(-6, 0)} $.

Eccentricity: $e = \frac{c}{a} = \frac{10}{6} = \mathbf{\frac{5}{3}}$.

Length of latus rectum: $ \frac{2b^2}{a} = \frac{2 \times 64}{6} = \frac{128}{6} = \mathbf{\frac{64}{3}}$ units.

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 and 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 into the standard form $ \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 $, we need to divide both sides of the equation by 36.

$ \frac{5y^2}{36} - \frac{9x^2}{36} = \frac{36}{36} $

Rewrite the fractions:

$ \frac{y^2}{\frac{36}{5}} - \frac{\cancel{9}^{1}x^2}{\cancel{36}_{4}} = 1 $

This gives the standard form of the hyperbola equation:

This equation is of the standard form $ \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 $ (since the $y^2$ term is positive, indicating the transverse axis is along the y-axis).

Comparing equation (1) with the standard form, we have:

From (2) and (3), we get $a = \sqrt{\frac{36}{5}} = \frac{6}{\sqrt{5}} = \frac{6\sqrt{5}}{5}$ and $b = \sqrt{4} = 2$.

Since the transverse axis is along the y-axis, the coordinates of the foci are $ (0, \pm c) $ and the coordinates of the vertices are $ (0, \pm a) $, where $c^2 = a^2 + b^2$.

Let's find the value of $c$:

Substitute the values of $a^2$ and $b^2$ from (2) and (3) into (4):

$c^2 = \frac{36}{5} + 4$

$c^2 = \frac{36}{5} + \frac{4 \times 5}{5} $

$c^2 = \frac{36 + 20}{5} $

$c^2 = \frac{56}{5} $

$c = \sqrt{\frac{56}{5}} = \frac{\sqrt{56}}{\sqrt{5}} = \frac{\sqrt{4 \times 14}}{\sqrt{5}} = \frac{2\sqrt{14}}{\sqrt{5}} = \frac{2\sqrt{14}\sqrt{5}}{5} = \frac{2\sqrt{70}}{5} $

Now we can find the required properties:

Coordinates of the foci: $ (0, \pm c) = \mathbf{\left(0, \pm \frac{2\sqrt{70}}{5}\right)} $.

The foci are at $ \left( 0, \frac{2\sqrt{70}}{5} \right) $ and $ \left( 0, -\frac{2\sqrt{70}}{5} \right) $.

Coordinates of the vertices: $ (0, \pm a) = \mathbf{\left(0, \pm \frac{6\sqrt{5}}{5}\right)} $.

The vertices are at $ \left( 0, \frac{6\sqrt{5}}{5} \right) $ and $ \left( 0, -\frac{6\sqrt{5}}{5} \right) $.

Eccentricity: $e = \frac{c}{a} = \frac{\frac{2\sqrt{70}}{5}}{\frac{6\sqrt{5}}{5}} = \frac{2\sqrt{70}}{5} \times \frac{5}{6\sqrt{5}} = \frac{2\sqrt{70}}{6\sqrt{5}} = \frac{\sqrt{14 \times 5}}{3\sqrt{5}} = \frac{\sqrt{14}\cancel{\sqrt{5}}}{3\cancel{\sqrt{5}}} = \frac{\sqrt{14}}{3} = \mathbf{\frac{\sqrt{14}}{3}}$.

Length of latus rectum: $ \frac{2b^2}{a} = \frac{2 \times 4}{\frac{6\sqrt{5}}{5}} = \frac{8}{\frac{6\sqrt{5}}{5}} = 8 \times \frac{5}{6\sqrt{5}} = \frac{40}{6\sqrt{5}} = \frac{20}{3\sqrt{5}} = \frac{20\sqrt{5}}{3 \times 5} = \frac{20\sqrt{5}}{15} = \mathbf{\frac{4\sqrt{5}}{3}}$ units.

Summary of Results:

Foci: $ \left(0, \pm \frac{2\sqrt{70}}{5}\right) $

Vertices: $ \left(0, \pm \frac{6\sqrt{5}}{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 and 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 into the standard form $ \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 $, we need to divide both sides of the equation by 784.

$ \frac{49y^2}{784} - \frac{16x^2}{784} = \frac{784}{784} $

Simplify the fractions:

$ \frac{\cancel{49}^{1}y^2}{\cancel{784}_{16}} - \frac{\cancel{16}^{1}x^2}{\cancel{784}_{49}} = 1 $

(Note: $ 784 \div 49 = 16 $ and $ 784 \div 16 = 49 $)

This gives the standard form of the hyperbola equation:

This equation is of the standard form $ \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 $ (since the $y^2$ term is positive, indicating the transverse axis is along the y-axis).

Comparing equation (1) with the standard form, we have:

From (2) and (3), we get $a = \sqrt{16} = 4$ and $b = \sqrt{49} = 7$.

Since the transverse axis is along the y-axis, the coordinates of the foci are $ (0, \pm c) $ and the coordinates of the vertices are $ (0, \pm a) $, where $c^2 = a^2 + b^2$.

Let's find the value of $c$:

Substitute the values of $a^2$ and $b^2$ from (2) and (3) into (4):

$c^2 = 16 + 49$

$c^2 = 65$

$c = \sqrt{65} $

Now we can find the required properties:

Coordinates of the foci: $ (0, \pm c) = \mathbf{(0, \pm \sqrt{65})} $.

The foci are at $ (0, \sqrt{65}) $ and $ (0, -\sqrt{65}) $.

Coordinates of the vertices: $ (0, \pm a) = \mathbf{(0, \pm 4)} $.

The vertices are at $ (0, 4) $ and $ (0, -4) $.

Eccentricity: $e = \frac{c}{a} = \frac{\sqrt{65}}{4} = \mathbf{\frac{\sqrt{65}}{4}}$.

Length of latus rectum: $ \frac{2b^2}{a} = \frac{2 \times 49}{4} = \frac{98}{4} = \frac{49}{2} = \mathbf{24.5}$ units.

Summary of Results:

Foci: $ (0, \pm \sqrt{65}) $

Vertices: $ (0, \pm 4) $

Eccentricity: $ \frac{\sqrt{65}}{4} $

Length of Latus Rectum: $ \frac{49}{2} $ or $ 24.5 $

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:

Since the vertices are $(\pm 2, 0)$ and the foci are $(\pm 3, 0)$, the center of the hyperbola is at the origin $ (0, 0) $.

Also, the vertices and foci lie on the x-axis. This indicates that the transverse axis of the hyperbola is along the x-axis.

The standard form of the equation of a hyperbola with the center at the origin and transverse axis along the x-axis is:

where $a$ is the distance from the center to each vertex along the transverse axis, and $b$ is the distance from the center to each end of the conjugate axis.

The coordinates of the vertices of such a hyperbola are $(\pm a, 0)$.

Comparing the given vertices $(\pm 2, 0)$ with $(\pm a, 0)$, we get:

So, $a^2 = 2^2 = 4$.

The coordinates of the foci of such a hyperbola are $(\pm c, 0)$, where $c$ is the distance from the center to each focus.

Comparing the given foci $(\pm 3, 0)$ with $(\pm c, 0)$, we get:

So, $c^2 = 3^2 = 9$.

We use the relationship between $a$, $b$, and $c$ for a hyperbola, which is $c^2 = a^2 + b^2$.

We can rearrange this to find $b^2$:

Substitute the values of $c^2 = 9$ (from 3) and $a^2 = 4$ (from 2) into (4):

$b^2 = 9 - 4$

$b^2 = 5$

Now, substitute the values of $a^2 = 4$ and $b^2 = 5$ into the standard equation of the hyperbola (1):

Thus, the equation of the hyperbola is $ \mathbf{\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:

Since the vertices are $(0, \pm 5)$ and the foci are $(0, \pm 8)$, the center of the hyperbola is at the origin $ (0, 0) $.

Also, the vertices and foci lie on the y-axis. This indicates that the transverse axis of the hyperbola is along the y-axis.

The standard form of the equation of a hyperbola with the center at the origin and transverse axis along the y-axis is:

where $a$ is the distance from the center to each vertex along the transverse axis, and $b$ is the distance from the center to each end of the conjugate axis.

The coordinates of the vertices of such a hyperbola are $(0, \pm a)$.

Comparing the given vertices $(0, \pm 5)$ with $(0, \pm a)$, we get:

So, $a^2 = 5^2 = 25$.

The coordinates of the foci of such a hyperbola are $(0, \pm c)$, where $c$ is the distance from the center to each focus.

Comparing the given foci $(0, \pm 8)$ with $(0, \pm c)$, we get:

So, $c^2 = 8^2 = 64$.

We use the relationship between $a$, $b$, and $c$ for a hyperbola, which is $c^2 = a^2 + b^2$.

We can rearrange this to find $b^2$:

Substitute the values of $c^2 = 64$ (from 3) and $a^2 = 25$ (from 2) into (4):

$b^2 = 64 - 25$

$b^2 = 39$

Now, substitute the values of $a^2 = 25$ and $b^2 = 39$ into the standard equation of the hyperbola (1):

Thus, the equation of the hyperbola is $ \mathbf{\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:

Since the vertices are $(0, \pm 3)$ and the foci are $(0, \pm 5)$, the center of the hyperbola is at the origin $ (0, 0) $.

Also, the vertices and foci lie on the y-axis. This indicates that the transverse axis of the hyperbola is along the y-axis.

The standard form of the equation of a hyperbola with the center at the origin and transverse axis along the y-axis is:

where $a$ is the distance from the center to each vertex along the transverse axis, and $b$ is the distance from the center to each end of the conjugate axis.

The coordinates of the vertices of such a hyperbola are $(0, \pm a)$.

Comparing the given vertices $(0, \pm 3)$ with $(0, \pm a)$, we get:

So, $a^2 = 3^2 = 9$.

The coordinates of the foci of such a hyperbola are $(0, \pm c)$, where $c$ is the distance from the center to each focus.

Comparing the given foci $(0, \pm 5)$ with $(0, \pm c)$, we get:

So, $c^2 = 5^2 = 25$.

We use the relationship between $a$, $b$, and $c$ for a hyperbola, which is $c^2 = a^2 + b^2$.

We can rearrange this to find $b^2$:

Substitute the values of $c^2 = 25$ (from 3) and $a^2 = 9$ (from 2) into (4):

$b^2 = 25 - 9$

$b^2 = 16$

Now, substitute the values of $a^2 = 9$ and $b^2 = 16$ into the standard equation of the hyperbola (1):

Thus, the equation of the hyperbola is $ \mathbf{\frac{y^2}{9} - \frac{x^2}{16} = 1} $.

Given:

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

The length of the transverse axis is 8.

To Find:

The equation of the hyperbola.

Solution:

Since the foci are located at $(\pm 5, 0)$, they lie on the x-axis.

This means the center of the hyperbola is at the origin $(0, 0)$ and the transverse axis is along the x-axis.

The standard form of the equation of a hyperbola with the center at the origin and transverse axis along the x-axis is:

where $a$ is the distance from the center to each vertex along the transverse axis, and $b$ is the distance from the center to each end of the conjugate axis.

The coordinates of the foci of such a hyperbola are $(\pm c, 0)$, where $c$ is the distance from the center to each focus.

Comparing the given foci $(\pm 5, 0)$ with $(\pm c, 0)$, we get:

So, $c^2 = 5^2 = 25$.

The length of the transverse axis is given as 8. For a hyperbola with the transverse axis along the x-axis, the length of the transverse axis is $2a$.

So, we have:

Dividing by 2, we get:

So, $a^2 = 4^2 = 16$.

We use the relationship between $a$, $b$, and $c$ for a hyperbola, which is $c^2 = a^2 + b^2$.

We can rearrange this to find $b^2$:

Substitute the values of $c^2 = 25$ (from 2) and $a^2 = 16$ (from 3) into (4):

$b^2 = 25 - 16$

$b^2 = 9$

Now, substitute the values of $a^2 = 16$ and $b^2 = 9$ into the standard equation of the hyperbola (1):

Thus, the equation of the hyperbola is $ \mathbf{\frac{x^2}{16} - \frac{y^2}{9} = 1} $.

Given:

The coordinates of the foci are $(0, \pm 13)$.

The length of the conjugate axis is 24.

To Find:

The equation of the hyperbola.

Solution:

Since the foci are located at $(0, \pm 13)$, they lie on the y-axis.

This means the center of the hyperbola is at the origin $(0, 0)$ and the transverse axis is along the y-axis.

The standard form of the equation of a hyperbola with the center at the origin and transverse axis along the y-axis is:

where $a$ is the distance from the center to each vertex along the transverse axis, and $b$ is the distance from the center to each end of the conjugate axis.

The coordinates of the foci of such a hyperbola are $(0, \pm c)$, where $c$ is the distance from the center to each focus.

Comparing the given foci $(0, \pm 13)$ with $(0, \pm c)$, we get:

So, $c^2 = 13^2 = 169$.

The length of the conjugate axis is given as 24. For a hyperbola, the length of the conjugate axis is $2b$.

So, we have:

Dividing by 2, we get:

So, $b^2 = 12^2 = 144$.

We use the relationship between $a$, $b$, and $c$ for a hyperbola, which is $c^2 = a^2 + b^2$.

We can rearrange this to find $a^2$:

Substitute the values of $c^2 = 169$ (from 2) and $b^2 = 144$ (from 3) into (4):

$a^2 = 169 - 144$

$a^2 = 25$

So, $a = \sqrt{25} = 5$.

Now, substitute the values of $a^2 = 25$ and $b^2 = 144$ into the standard equation of the hyperbola (1):

Thus, the equation of the hyperbola is $ \mathbf{\frac{y^2}{25} - \frac{x^2}{144} = 1} $.

Given:

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

The length of the latus rectum is 8.

To Find:

The equation of the hyperbola.

Solution:

Since the foci are located at $ (\pm 3\sqrt{5}, 0) $, they lie on the x-axis.

This means the center of the hyperbola is at the origin $(0, 0)$ and the transverse axis is along the x-axis.

The standard form of the equation of a hyperbola with the center at the origin and transverse axis along the x-axis is:

where $a$ is the distance from the center to each vertex along the transverse axis, and $b$ is the distance from the center to each end of the conjugate axis.

The coordinates of the foci of such a hyperbola are $(\pm c, 0)$, where $c$ is the distance from the center to each focus.

Comparing the given foci $ (\pm 3\sqrt{5}, 0) $ with $(\pm c, 0)$, we get:

So, $c^2 = (3\sqrt{5})^2 = 3^2 \times (\sqrt{5})^2 = 9 \times 5 = 45$.

The length of the latus rectum of a hyperbola with the transverse axis along the x-axis is given by the formula $ \frac{2b^2}{a} $. We are given that this length is 8.

From equation (3), we can express $2b^2$ in terms of $a$:

$2b^2 = 8a$

We use the relationship between $a$, $b$, and $c$ for a hyperbola, which is $c^2 = a^2 + b^2$.

Substitute the value of $c^2 = 45$ (from 2) and $b^2 = 4a$ (from 4) into (5):

$45 = a^2 + 4a$

Rearrange the equation into a quadratic equation in $a$:

$a^2 + 4a - 45 = 0$

We can solve this quadratic equation for $a$ by factoring. We look for two numbers that multiply to -45 and add up to 4. These numbers are 9 and -5.

$a^2 + 9a - 5a - 45 = 0$

$a(a + 9) - 5(a + 9) = 0$

$(a - 5)(a + 9) = 0$

This gives two possible values for $a$: $a = 5$ or $a = -9$.

Since $a$ represents a distance, it must be positive. Therefore, we take $a = 5$.

So, $a^2 = 5^2 = 25$.

Now, substitute the value of $a = 5$ into equation (4) to find $b^2$:

$b^2 = 4 \times 5$

Substitute the values of $a^2 = 25$ (from 6) and $b^2 = 20$ (from 7) into the standard equation of the hyperbola (1):

Thus, the equation of the hyperbola is $ \mathbf{\frac{x^2}{25} - \frac{y^2}{20} = 1} $.

Given:

The coordinates of the foci are $(\pm 4, 0)$.

The length of the latus rectum is 12.

To Find:

The equation of the hyperbola.

Solution:

Since the foci are located at $(\pm 4, 0)$, they lie on the x-axis.

This means the center of the hyperbola is at the origin $(0, 0)$ and the transverse axis is along the x-axis.

The standard form of the equation of a hyperbola with the center at the origin and transverse axis along the x-axis is:

where $a$ is the distance from the center to each vertex along the transverse axis, and $b$ is the distance from the center to each end of the conjugate axis.

The coordinates of the foci of such a hyperbola are $(\pm c, 0)$, where $c$ is the distance from the center to each focus.

Comparing the given foci $(\pm 4, 0)$ with $(\pm c, 0)$, we get:

So, $c^2 = 4^2 = 16$.

The length of the latus rectum of a hyperbola with the transverse axis along the x-axis is given by the formula $ \frac{2b^2}{a} $. We are given that this length is 12.

From equation (3), we can express $2b^2$ in terms of $a$:

$2b^2 = 12a$

We use the relationship between $a$, $b$, and $c$ for a hyperbola, which is $c^2 = a^2 + b^2$.

Substitute the value of $c^2 = 16$ (from 2) and $b^2 = 6a$ (from 4) into (5):

$16 = a^2 + 6a$

Rearrange the equation into a quadratic equation in $a$:

$a^2 + 6a - 16 = 0$

We can solve this quadratic equation for $a$ by factoring. We look for two numbers that multiply to -16 and add up to 6. These numbers are 8 and -2.

$a^2 + 8a - 2a - 16 = 0$

$a(a + 8) - 2(a + 8) = 0$

$(a - 2)(a + 8) = 0$

This gives two possible values for $a$: $a = 2$ or $a = -8$.

Since $a$ represents a distance, it must be positive. Therefore, we take $a = 2$.

So, $a^2 = 2^2 = 4$.

Now, substitute the value of $a = 2$ into equation (4) to find $b^2$:

$b^2 = 6 \times 2$

Substitute the values of $a^2 = 4$ (from 6) and $b^2 = 12$ (from 7) into the standard equation of the hyperbola (1):

Thus, the equation of the hyperbola is $ \mathbf{\frac{x^2}{4} - \frac{y^2}{12} = 1} $.

Given:

The coordinates of the vertices are $(\pm 7, 0)$.

The eccentricity $ e = \frac{4}{3} $.

To Find:

The equation of the hyperbola.

Solution:

Since the vertices are located at $(\pm 7, 0)$, they lie on the x-axis.

This means the center of the hyperbola is at the origin $(0, 0)$ and the transverse axis is along the x-axis.

The standard form of the equation of a hyperbola with the center at the origin and transverse axis along the x-axis is:

where $a$ is the distance from the center to each vertex along the transverse axis, and $b$ is the distance from the center to each end of the conjugate axis.

The coordinates of the vertices of such a hyperbola are $(\pm a, 0)$.

Comparing the given vertices $(\pm 7, 0)$ with $(\pm a, 0)$, we get:

So, $a^2 = 7^2 = 49$.

The eccentricity $e$ of a hyperbola is given by the formula $ e = \frac{c}{a} $, where $c$ is the distance from the center to each focus.

We are given the eccentricity $e = \frac{4}{3}$.

Substitute the value of $a = 7$ from (2) into (3):

$ \frac{c}{7} = \frac{4}{3} $

Multiply both sides by 7:

$ c = 7 \times \frac{4}{3} = \frac{28}{3} $

So, $c^2 = \left(\frac{28}{3}\right)^2 = \frac{784}{9}$.

We use the relationship between $a$, $b$, and $c$ for a hyperbola, which is $c^2 = a^2 + b^2$.

We can rearrange this to find $b^2$:

Substitute the values of $c^2 = \frac{784}{9}$ and $a^2 = 49 = \frac{49 \times 9}{9} = \frac{441}{9}$ (from 2) into (5):

$b^2 = \frac{784}{9} - \frac{441}{9} $

$b^2 = \frac{784 - 441}{9} $

$b^2 = \frac{343}{9} $

Substitute the values of $a^2 = 49$ (from 2) and $b^2 = \frac{343}{9}$ (from 6) into the standard equation of the hyperbola (1):

Simplify the fraction in the denominator of the second term:

$ \frac{x^2}{49} - \frac{9y^2}{343} = 1 $

Thus, the equation of the hyperbola is $ \mathbf{\frac{x^2}{49} - \frac{9y^2}{343} = 1} $.

Given:

The coordinates of the foci are $(0, \pm \sqrt{10})$.

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

To Find:

The equation of the hyperbola.

Solution:

Since the foci are located at $(0, \pm \sqrt{10})$, they lie on the y-axis.

This means the center of the hyperbola is at the origin $(0, 0)$ and the transverse axis is along the y-axis.

The standard form of the equation of a hyperbola with the center at the origin and transverse axis along the y-axis is:

where $a$ is the distance from the center to each vertex along the transverse axis, and $b$ is the distance from the center to each end of the conjugate axis. For a hyperbola with the transverse axis along the y-axis, the vertices are $(0, \pm a)$ and the foci are $(0, \pm c)$, and the relation between $a, b, c$ is $c^2 = a^2 + b^2$.

The coordinates of the foci are $(0, \pm c)$. Comparing the given foci $(0, \pm \sqrt{10})$ with $(0, \pm c)$, we get:

So, $c^2 = (\sqrt{10})^2 = 10$.

We use the relationship $c^2 = a^2 + b^2$. Substitute $c^2 = 10$:

From equation (3), we can express $b^2$ in terms of $a^2$:

The hyperbola passes through the point $ (2, 3) $. Substituting $x=2$ and $y=3$ into equation (1), we get:

This simplifies to:

Substitute the expression for $b^2$ from equation (4) into equation (6):

$ \frac{9}{a^2} - \frac{4}{10 - a^2} = 1 $

To eliminate the denominators, multiply both sides by $ a^2(10 - a^2) $ (assuming $a^2 \neq 0$ and $a^2 \neq 10$).

$ 9(10 - a^2) - 4a^2 = a^2(10 - a^2) $

$ 90 - 9a^2 - 4a^2 = 10a^2 - a^4 $

$ 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 $ Z = a^2 $. The equation becomes a quadratic equation in $Z$:

$ Z^2 - 23Z + 90 = 0 $

Factor the quadratic equation:

$(Z - 18)(Z - 5) = 0 $

This gives two possible values for $Z = a^2$: $ a^2 = 18 $ or $ a^2 = 5 $.

For a hyperbola with the transverse axis along the y-axis, the vertices are $(0, \pm a)$ and the foci are $(0, \pm c)$. The vertices lie between the center and the foci, so $a < c$.

From equation (2), $c = \sqrt{10}$, so $c^2 = 10$.

We must have $ a < c $, which means $ a^2 < c^2 $.

Since $ c^2 = 10 $, we need $ a^2 < 10 $.

Out of the two possible values for $ a^2 $, $a^2 = 18$ and $a^2 = 5$, only $ a^2 = 5 $ satisfies the condition $ a^2 < 10 $.

Now find $b^2$ using equation (4):

$b^2 = 10 - a^2 = 10 - 5 $

Substitute the values of $a^2 = 5$ (from 7) and $b^2 = 5$ (from 8) into the standard equation of the hyperbola (1):

Thus, the equation of the hyperbola is $ \mathbf{\frac{y^2}{5} - \frac{x^2}{5} = 1} $. This is a rectangular hyperbola since $a=b=\sqrt{5}$.

Given:

The parabolic mirror has its vertex at the origin (0, 0).

The focus (F) is at a distance of 5 cm from the vertex.

The mirror is 45 cm deep.

To Find:

The distance AB as shown in Fig 10.31.

Solution:

From the figure, the vertex of the parabolic mirror is at the origin $(0,0)$, and the focus lies on the positive x-axis.

The standard equation of a parabola with vertex at the origin and focus on the positive x-axis is $y^2 = 4ax$.

Here, $a$ is the distance of the focus from the vertex.

We are given that the distance of the focus from the vertex is 5 cm.

Substitute the value of $a$ into the standard equation of the parabola:

So, the equation of the parabola is $ y^2 = 20x $.

The depth of the mirror is given as 45 cm. This corresponds to the x-coordinate of the points A and B.

To find the y-coordinates of the points A and B, substitute $x = 45$ into equation (1):

$y^2 = 20 \times 45$

$y^2 = 900$

Take the square root of both sides to find $y$:

$y = \pm \sqrt{900}$

$y = \pm 30$

The points A and B lie on the parabola at the depth of 45 cm. Their coordinates are $(45, 30)$ and $(45, -30)$.

The distance AB is the vertical distance between these two points. The y-coordinate of point A is 30, and the y-coordinate of point B is -30.

Distance AB = $ |y_A - y_B| = |30 - (-30)| = |30 + 30| = |60| = 60 $ cm.

Alternatively, since the parabola is symmetric about the x-axis, the distance AB is twice the absolute value of the y-coordinate of A (or B).

Distance AB = $ 2 \times |y_A| = 2 \times |30| = 2 \times 30 = 60 $ cm.

Therefore, the distance AB is 60 cm.

Given:

To Find:

The distance from the center where the deflection is 1 cm.

Solution:

Let's set up a coordinate system to model the shape of the deflected beam, which is a parabola.

Place the origin $(0,0)$ at the lowest point of the beam, which is at the center. Let the y-axis point upwards along the axis of the parabola. The x-axis is horizontal.

The standard equation of a parabola with its vertex at the origin $(0,0)$ and opening upwards is $x^2 = 4ay$, where $a > 0$ is the distance from the vertex to the focus.

The supports are 12 metres apart, centered around the origin. So, the horizontal distance from the center (origin) to each support is $\frac{12}{2} = 6$ metres.

The maximum deflection at the center is 3 cm. This means the lowest point of the beam (the origin in our setup) is 3 cm below the level of the supports. Therefore, the supports are located at a height of 3 cm above the origin.

To use consistent units, let's convert all measurements to metres. 1 cm = 0.01 metres.

Distance between supports = 12 metres.

Horizontal distance to supports = 6 metres.

Maximum deflection = 3 cm = 0.03 metres.

In our coordinate system, the supports are located at $x = \pm 6$ metres, and their y-coordinate is 0.03 metres (since they are 0.03 metres above the origin).

So, the parabola passes through the points $(\pm 6, 0.03)$.

Substitute the coordinates of one of the support points, say $(6, 0.03)$, into the parabola equation $x^2 = 4ay$:

Solve for $a$:

So, the equation of the parabola is $x^2 = 4(300)y$, which simplifies to:

Here, $x$ is the horizontal distance from the center (in metres) and $y$ is the height above the lowest point of the beam (in metres).

We need to find the distance from the center ($x$) where the deflection is 1 cm. Deflection is the vertical distance from the level of the supports down to the beam.

The level of the supports is at $y = 0.03$ metres in our coordinate system.

A deflection of 1 cm = 0.01 metres means the vertical distance from the support level (0.03 m) downwards is 0.01 m. The y-coordinate of such a point on the parabola is $0.03 \text{ m} - 0.01 \text{ m} = 0.02$ metres.

So, we need to find the value of $x$ when $y = 0.02$ metres.

Substitute $y = 0.02$ into the equation of the parabola (1):

Take the square root of both sides to find $x$:

Simplify the square root:

So, $x = \pm 2\sqrt{6}$ metres.

The distance from the center (origin) is the absolute value of the x-coordinate.

Distance $= |x| = | \pm 2\sqrt{6} | = 2\sqrt{6}$ metres.

Answer:

The distance from the center where the deflection is 1 cm is $\mathbf{2\sqrt{6}}$ metres.

Given:

To Show:

The locus of point P(x, y) is an ellipse.

Solution:

Let the coordinates of the end point A on the x-axis be $(a, 0)$, where $a$ is the x-intercept. Since A is on the x-axis, its y-coordinate is 0.

Let the coordinates of the end point B on the y-axis be $(0, b)$, where $b$ is the y-intercept. Since B is on the y-axis, its x-coordinate is 0.

The length of the rod AB is given as 15 cm. We can use the distance formula to find the distance between points A $(a, 0)$ and B $(0, b)$:

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

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

Since $AB = 15$ cm, we have:

Squaring both sides of equation (1):

Point P(x, y) lies on the rod AB such that AP = 6 cm. The total length of the rod is AB = 15 cm.

The length of the segment PB is $PB = AB - AP = 15 - 6 = 9$ cm.

This means that the point P divides the line segment AB in the ratio $AP : PB = 6 : 9$.

Simplify the ratio: $6 : 9 = \frac{6}{3} : \frac{9}{3} = 2 : 3$.

So, point P divides the segment AB internally in the ratio $2 : 3$. Here, the ratio is $m:n = 2:3$, where A is the first point and B is the second point for the section formula.

We can use the section formula to find the coordinates of point P(x, y) that divides the line segment joining $A(x_1, y_1) = (a, 0)$ and $B(x_2, y_2) = (0, b)$ in the ratio $m:n = 2:3$.

The coordinates of P are given by:

$x = \frac{n x_1 + m x_2}{m + n}$

$y = \frac{n y_1 + m y_2}{m + n}$

Substitute the coordinates of A and B and the ratio m:n = 2:3:

For the x-coordinate of P:

$x = \frac{3 \times a + 2 \times 0}{2 + 3} = \frac{3a}{5}$

For the y-coordinate of P:

$y = \frac{3 \times 0 + 2 \times b}{2 + 3} = \frac{2b}{5}$

Now, we need to eliminate the variables $a$ and $b$ from equation (2) using equations (3) and (4). From equation (3), solve for $a$ in terms of $x$:

$5x = 3a \implies a = \frac{5x}{3}$

From equation (4), solve for $b$ in terms of $y$:

$5y = 2b \implies b = \frac{5y}{2}$

Substitute the expressions for $a$ from (5) and $b$ from (6) into equation (2):

Square the terms in the parentheses:

To obtain the standard form of the ellipse equation ($\frac{x^2}{A^2} + \frac{y^2}{B^2} = 1$), divide both sides of the equation by 225:

Simplify the denominators by cancelling the common factor 25:

This equation is of the form $\frac{x^2}{A^2} + \frac{y^2}{B^2} = 1$, where $A^2 = 81$ and $B^2 = 36$. This is the standard equation of an ellipse centered at the origin.

Since the equation derived for the coordinates $(x, y)$ of point P is the standard equation of an ellipse, the locus of point P is an ellipse.

Conclusion:

The derived equation for the locus of point P is $\mathbf{\frac{x^2}{81} + \frac{y^2}{36} = 1}$, which is the equation of an ellipse. Therefore, the locus of P is an ellipse.

Given:

To Find:

The position of the focus of the parabolic reflector.

Solution:

We can model the parabolic reflector using a standard equation of a parabola. Let's place the vertex of the parabola at the origin $(0,0)$ and the axis of symmetry along the x-axis. Since it's a reflector, it typically opens towards the focus.

If the parabola opens to the right, its equation is $y^2 = 4ax$. If it opens to the left, it's $y^2 = -4ax$. If it opens upwards, $x^2 = 4ay$, and downwards, $x^2 = -4ay$.

A parabolic reflector uses the property that rays parallel to the axis of symmetry are reflected towards the focus (or rays from the focus are reflected parallel to the axis). Given the shape described by diameter and depth, placing the vertex at the origin and the axis along either the x or y-axis is appropriate.

Let's assume the vertex is at $(0,0)$ and the axis of symmetry is along the positive x-axis. The parabola opens to the right. The equation is $y^2 = 4ax$. The focus is at $(a, 0)$.

The depth of the reflector is 5 cm. This means the furthest point on the parabola along the x-axis (from the vertex) is at $x = 5$.

The diameter is 20 cm. This is the length of the segment perpendicular to the axis of symmetry at $x=5$. The endpoints of this segment lie on the parabola. Let these points be $(5, y')$ and $(5, -y')$. The distance between these points is $y' - (-y') = 2y'$.

Given the diameter is 20 cm, $2y' = 20$, so $y' = 10$.

Thus, the points $(5, 10)$ and $(5, -10)$ lie on the parabola $y^2 = 4ax$.

Substitute the coordinates of the point $(5, 10)$ into the equation $y^2 = 4ax$:

Solve for $a$:

The value of $a$ is the distance of the focus from the vertex along the axis of symmetry.

Since the vertex is at $(0,0)$ and the axis is the positive x-axis, the focus is at $(a, 0)$.

Focus coordinates = $(5, 0)$.

The focus is located at a distance of 5 cm from the vertex along the axis of symmetry.

Answer:

The focus of the parabolic reflector is located at $\mathbf{(5, 0)}$ if the vertex is at $(0,0)$ and it opens along the positive x-axis. The distance of the focus from the vertex is $\mathbf{5}$ cm.

Given:

To Find:

The width of the arch at a height 2 m from the vertex of the parabola.

Solution:

Let's model the parabolic arch using a coordinate system. We can place the vertex of the parabola at the origin $(0,0)$ and the axis of symmetry along the y-axis. Since it's an arch with the base below the vertex, the parabola opens downwards.

The standard equation of a parabola with vertex at $(0,0)$, axis along the y-axis, and opening downwards is $x^2 = -4ay$, where $a > 0$ is the distance from the vertex to the focus.

The height of the arch is 10 m. This means the base of the arch is 10 m below the vertex. So, the y-coordinate of the base is $y = -10$ m.

The width of the arch at the base is 5 m. The base lies on the line $y = -10$. The points where the parabola intersects this line are at $x = \pm \frac{5}{2} = \pm 2.5$ m.

So, the points $(\pm 2.5, -10)$ lie on the parabola $x^2 = -4ay$.

Substitute the coordinates of the point $(2.5, -10)$ into the equation of the parabola:

Solve for $a$:

Simplify the fraction:

Substitute the value of $a$ into the standard equation of the parabola:

We need to find the width of the arch at a height 2 m from the vertex. Since the vertex is at $(0,0)$ and the axis is vertical (y-axis), a distance of 2 m downwards from the vertex corresponds to the y-coordinate $y = -2$ m.

Substitute $y = -2$ into the equation of the parabola (2):

Take the square root of both sides to find the corresponding x-coordinates:

The points on the parabola at this height are $ \left(\frac{\sqrt{5}}{2}, -2\right) $ and $ \left(-\frac{\sqrt{5}}{2}, -2\right) $.

The width of the arch at this height is the horizontal distance between these two points.

Width = $ \left| \frac{\sqrt{5}}{2} - \left(-\frac{\sqrt{5}}{2}\right) \right| = \left| \frac{\sqrt{5}}{2} + \frac{\sqrt{5}}{2} \right| = \left| \frac{2\sqrt{5}}{2} \right| = |\sqrt{5}| = \sqrt{5} $ metres.

Answer:

The width of the arch 2 m from the vertex is $\mathbf{\sqrt{5}}$ metres.

Given:

To Find:

The length of a supporting wire attached to the roadway 18 m from the middle.

Solution:

Let's set up a coordinate system. We place the origin $(0,0)$ at the middle of the roadway. The roadway is horizontal, so it lies along the x-axis.

The length of the roadway is 100 m, centered at the origin. So the roadway extends from $x = -50$ to $x = 50$. The ends of the roadway are at $x = \pm 50$.

The cable hangs above the roadway. The shortest vertical supporting wire is at the middle of the roadway ($x=0$). Its length is 6 m. This wire connects the point $(0,0)$ on the roadway to the lowest point of the cable, which is the vertex of the parabola.

Since the vertex is 6 m above the middle of the roadway, its coordinates are $(0, 6)$.

The axis of the parabolic cable is vertical (along the y-axis), and the parabola opens upwards. The standard equation of a parabola with vertex at $(h, k)$ is $(x-h)^2 = 4a(y-k)$.

With vertex at $(0, 6)$, the equation is $(x-0)^2 = 4a(y-6)$, which simplifies to:

The longest wires are at the ends of the roadway, $x = \pm 50$. Their length is 30 m. The roadway is at $y=0$. A wire of length 30 m at $x=50$ connects the point $(50, 0)$ on the roadway to a point on the cable. The y-coordinate of this point on the cable must be 30 (height above the roadway).

So, the point $(50, 30)$ lies on the parabolic cable.

Substitute the coordinates of the point $(50, 30)$ into the equation of the parabola (1):

Solve for $a$:

Simplify the fraction:

Substitute the value of $a$ back into equation (1) to get the equation of the parabola representing the cable:

We need to find the length of the supporting wire attached to the roadway 18 m from the middle. This corresponds to a horizontal distance of $x = 18$ m from the origin (middle).

The length of the wire at horizontal position $x$ is the y-coordinate of the cable at that x-value, since the roadway is at $y=0$. We can use equation (3) to find the y-coordinate when $x=18$.

Substitute $x = 18$ into equation (3):

Solve for $(y - 6)$:

Solve for $y$:

Find a common denominator to add the terms:

The y-coordinate represents the height of the cable above the roadway at $x=18$. This height is the length of the supporting wire.

Length of the supporting wire at 18 m from the middle = $\frac{5694}{625}$ metres.

Answer:

The length of the supporting wire attached to the roadway 18 m from the middle is $\mathbf{\frac{5694}{625}}$ metres.

Given:

To Find:

The height of the arch at a point 1.5 m from one end.

Solution:

Let's set up a coordinate system with the center of the base of the semi-ellipse at the origin $(0,0)$. The base lies on the x-axis.

The arch is 8 m wide at the base. This means the base extends from $x = -\frac{8}{2}$ to $x = \frac{8}{2}$, i.e., from $x = -4$ to $x = 4$. The ends of the base are at $(\pm 4, 0)$.

Since the base lies along the x-axis and the arch is a semi-ellipse extending upwards, the major axis of the full ellipse (if it were complete) would be along the x-axis, and the vertices would be $(\pm 4, 0)$. The length of the semi-major axis is $a = 4$.

So, $a^2 = 4^2 = 16$.

The height of the arch at the center is 2 m. The center of the base is the origin. The highest point of the semi-ellipse is at $(0, 2)$. This point is the upper endpoint of the semi-minor axis (since the major axis is horizontal).

The length of the semi-minor axis is $b = 2$.

So, $b^2 = 2^2 = 4$.

The standard equation of an ellipse centered at the origin with the major axis along the x-axis is $ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $.

Substitute the values of $a^2 = 16$ and $b^2 = 4$ into the standard equation:

Since it is a semi-ellipse forming an arch above the base, we consider the part of the ellipse where $y \ge 0$.

We need to find the height of the arch at a point 1.5 m from one end. Let's consider the end point on the right, which is at $x = 4$. A point 1.5 m from this end, along the base (x-axis), is at a horizontal distance from the origin $x = 4 - 1.5 = 2.5$ m.

We need to find the y-coordinate (height) of the ellipse at $x = 2.5$. Substitute $x = 2.5 = \frac{5}{2}$ into equation (3):

Subtract $ \frac{25}{64} $ from both sides:

Multiply both sides by 4:

Take the square root of both sides to find $y$. Since we are looking for the height of the arch above the base, we take the positive square root.

The height of the arch at a point 1.5 m from one end is $ \frac{\sqrt{39}}{4} $ metres.

Answer:

The height of the arch at a point 1.5 m from one end is $\mathbf{\frac{\sqrt{39}}{4}}$ metres.

Given:

To Find:

The equation of the locus of point P.

Solution:

Let the rod be AB, where end A is always on the x-axis and end B is always on the y-axis. The length of the rod is $AB = 12$ cm.

Let the coordinates of point A on the x-axis be $(a, 0)$, where $a$ is the x-intercept. Since A is on the x-axis, its y-coordinate is 0.

Let the coordinates of point B on the y-axis be $(0, b)$, where $b$ is the y-intercept. Since B is on the y-axis, its x-coordinate is 0.

Using the distance formula for the length of the rod AB:

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

Given $AB = 12$ cm:

Squaring both sides of equation (1):

Let P(x, y) be a point on the rod. Point P is 3 cm from the end in contact with the x-axis, which is point A. So, $AP = 3$ cm.

The length of the rod is $AB = 12$ cm. The length of the segment PB is $PB = AB - AP = 12 - 3 = 9$ cm.

Thus, point P divides the line segment AB internally in the ratio $AP : PB = 3 : 9 = 1 : 3$.

We can use the section formula to find the coordinates of P(x, y) that divides the line segment joining $A(x_1, y_1) = (a, 0)$ and $B(x_2, y_2) = (0, b)$ in the ratio $m:n = 1:3$.

The coordinates of P are:

$x = \frac{n x_1 + m x_2}{m + n} = \frac{3 \times a + 1 \times 0}{1 + 3} = \frac{3a}{4}$

$y = \frac{n y_1 + m y_2}{m + n} = \frac{3 \times 0 + 1 \times b}{1 + 3} = \frac{b}{4}$

Now, we express $a$ and $b$ in terms of $x$ and $y$ from equations (3) and (4):

From (3): $4x = 3a \implies a = \frac{4x}{3}$

From (4): $4y = b \implies b = 4y$

Substitute the expressions for $a$ from (5) and $b$ from (6) into equation (2):

Square the terms:

Divide the entire equation by 16 to simplify:

To get the standard form of the ellipse equation ($\frac{x^2}{A^2} + \frac{y^2}{B^2} = 1$), divide both sides by 9:

This is the standard equation of an ellipse centered at the origin with semi-major axis $A = \sqrt{81} = 9$ along the x-axis and semi-minor axis $B = \sqrt{9} = 3$ along the y-axis.

Answer:

The equation of the locus of point P is $\mathbf{\frac{x^2}{81} + \frac{y^2}{9} = 1}$.

Given:

To Find:

The area of the triangle formed by the vertex of the parabola and the ends of its latus rectum.

Solution:

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

This equation is of the standard form $ x^2 = 4ay $. Comparing the given equation with the standard form, we find the value of $a$:

Divide both sides by 4:

For a parabola of the form $x^2 = 4ay$, the vertex is at the origin $(0,0)$, and the axis of symmetry is the y-axis. Since $a=3 > 0$, the parabola opens upwards.

The vertex of the parabola (V) is at $ \mathbf{(0, 0)} $.

The latus rectum of this parabola is a line segment perpendicular to the axis of symmetry (y-axis), passing through the focus, and with endpoints on the parabola. The focus (F) is at $(0, a)$. With $a=3$, the focus is at $(0, 3)$. The equation of the line containing the latus rectum is $y = a$, so $y = 3$.

To find the endpoints of the latus rectum, substitute $y = 3$ into the equation of the parabola $x^2 = 12y$:

Take the square root of both sides:

The endpoints of the latus rectum are at $x = -6$ and $x = 6$ when $y=3$. Let these points be L and R.

The coordinates of the endpoints of the latus rectum are $ \mathbf{L = (-6, 3)} $ and $ \mathbf{R = (6, 3)} $.

We need to find the area of the triangle formed by the vertex $V(0, 0)$ and the endpoints of the latus rectum $L(-6, 3)$ and $R(6, 3)$.

We can find the area of this triangle using the formula for the area of a triangle with vertices $ (x_1, y_1), (x_2, y_2), $ and $ (x_3, y_3) $.

Area $= \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| $

Let $ (x_1, y_1) = V(0, 0) $, $ (x_2, y_2) = L(-6, 3) $, and $ (x_3, y_3) = R(6, 3) $.

Alternate Method (Base and Height):

The triangle VLR has vertices $V(0,0)$, $L(-6, 3)$, and $R(6, 3)$.

We can consider the base of the triangle to be the line segment LR. Since L and R have the same y-coordinate (3), the base LR is a horizontal segment. The length of the base is the distance between the x-coordinates: $|6 - (-6)| = |12| = 12$ units. This is also the length of the latus rectum $4a = 4(3) = 12$.

The height of the triangle is the perpendicular distance from the vertex V $(0, 0)$ to the line containing the base LR, which is the line $y = 3$. The distance from the origin $(0,0)$ to the horizontal line $y=3$ is the absolute value of the y-coordinate, which is $|3| = 3$ units.

Area of the triangle = $ \frac{1}{2} \times \text{base} \times \text{height} $

Both methods give the same result.

Answer:

The area of the triangle is $\mathbf{18}$ square units.

Given:

To Find:

The equation of the locus traced by the man.

Solution:

Let the two flag posts be fixed points, which we call foci. Let F1 and F2 be the locations of the two flag posts.

Let P be the position of the man at any point in time. The problem states that the sum of the distances from the man to the two flag posts is always 10 m. This means $PF1 + PF2 = 10$.

The definition of an ellipse is the locus of a point for which the sum of its distances from two fixed points (the foci) is a constant value.

Therefore, the path traced by the man is an ellipse, and the two flag posts F1 and F2 are the foci of this ellipse.

Let the distance between the foci be $2c$. Given the distance between the flag posts is 8 m:

Divide by 2:

The constant sum of the distances from any point on the ellipse to the foci is equal to the length of the major axis, which is $2a$.

Given that the sum of the distances ($PF1 + PF2$) is always 10 m:

Divide by 2:

So, $a^2 = 5^2 = 25$.

For an ellipse, the relationship between $a$, $b$ (semi-minor axis), and $c$ (distance from center to focus) is $c^2 = a^2 - b^2$. We can rearrange this to find $b^2$:

Substitute the values of $a^2 = 25$ (from 2) and $c^2 = 4^2 = 16$ (from 1) into equation (3):

So, $b = \sqrt{9} = 3$. Note that $a=5 > b=3$, which is required for $a$ to be the semi-major axis length.

Let's place the center of the ellipse at the origin $(0,0)$ and the foci on the x-axis at $(-c, 0)$ and $(c, 0)$, i.e., $(-4, 0)$ and $(4, 0)$. The major axis is along the x-axis.

The standard form of the equation of an ellipse centered at the origin with the major axis along the x-axis is:

Substitute the values of $a^2 = 25$ and $b^2 = 9$ into equation (4):

This is the equation of the locus traced by the man.

Answer:

The equation of the locus traced by the man is $\mathbf{\frac{x^2}{25} + \frac{y^2}{9} = 1}$.

Given:

To Find:

The length of the side of the equilateral triangle in terms of $a$.

Solution:

The equation of the parabola is $ y^2 = 4ax $. The vertex of this parabola is at the origin $(0,0)$. The axis of symmetry is the x-axis, and since the coefficient of $x$ is $4a$ (assuming $a>0$ from standard form context), the parabola opens to the right.

Let the vertices of the equilateral triangle be O, P, and Q. One vertex is at the vertex of the parabola, so let $O = (0, 0)$.

Since the triangle is inscribed in the parabola, the other two vertices P and Q must lie on the parabola $y^2 = 4ax$.

Because the parabola is symmetric about the x-axis and one vertex is on the axis of symmetry (at the origin), the other two vertices P and Q must be symmetric with respect to the x-axis. Let the coordinates of P be $(x_1, y_1)$. Then the coordinates of Q must be $(x_1, -y_1)$. Since O, P, Q form a triangle, $y_1$ cannot be 0.

Let $s$ be the length of the side of the equilateral triangle. The distance between any two vertices is $s$.

Distance OP = $s$. Using the distance formula between O(0,0) and P$(x_1, y_1)$:

Distance OQ = $s$. Using the distance formula between O(0,0) and Q$(x_1, -y_1)$:

Distance PQ = $s$. Using the distance formula between P$(x_1, y_1)$ and Q$(x_1, -y_1)$:

Since $y_1 \neq 0$, we can write $s = 2|y_1|$. Let's assume $y_1 > 0$ without loss of generality, so $s = 2y_1$. This gives $y_1 = \frac{s}{2}$.

From equation (2), substitute $y_1 = s/2$ into equation (1):

$s^2 = x_1^2 + \left(\frac{s}{2}\right)^2$

$s^2 = x_1^2 + \frac{s^2}{4}$

Subtract $\frac{s^2}{4}$ from both sides:

$s^2 - \frac{s^2}{4} = x_1^2$

Take the square root of both sides:

$ \sqrt{\frac{3s^2}{4}} = |x_1| $

$ \frac{\sqrt{3}|s|}{2} = |x_1| $

Since $s = 2y_1$ and $y_1 > 0$, $s > 0$. For the parabola $y^2 = 4ax$ with $a>0$ opening to the right, a point P$(x_1, y_1)$ with $y_1 > 0$ must have $x_1 > 0$. So $|s|=s$ and $|x_1|=x_1$.

The point P$(x_1, y_1)$ lies on the parabola $y^2 = 4ax$. Substitute $x_1$ and $y_1$ in terms of $s$ into the parabola equation $y_1^2 = 4ax_1$:

Since $s \neq 0$, we can divide both sides by $s$:

Multiply both sides by 4 to solve for $s$:

The length of the side of the equilateral triangle is $8\sqrt{3}a$.

Answer:

The length of the side of the triangle is $\mathbf{8\sqrt{3}a}$.