Chapter 11 Conic Sections

Given:

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

To Find:

The centre and the radius of the circle.

Solution:

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

The given equation of the circle is:

To find the centre and radius, we need to rewrite this equation in the standard form (1) by completing the square for the $x$ and $y$ terms.

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

Now, compare equation (2) with the standard form (1):

Thus, the centre of the circle is $(h, k) = (1, -2)$ and the radius is $r = \sqrt{13}$.

Answer:

The centre of the circle is $(1, -2)$.

The radius of the circle is $\sqrt{13}$.

Given:

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

To Find:

Coordinates of the focus, equation of the directrix, and length of the latus rectum.

Solution:

The given equation of the parabola is:

This equation is of the form $x^2 = -4ay$, which represents a parabola opening downwards with its vertex at the origin $(0,0)$.

Comparing the given equation with the standard form $x^2 = -4ay$, we have:

Dividing both sides by -4, we get:

For a parabola of the form $x^2 = -4ay$:

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

Substituting the value of $a$ from (i):

Focus = $(0, -2)$

2. The equation of the directrix is $y = a$.

Substituting the value of $a$ from (i):

Directrix equation is $y = 2$.

3. The length of the latus rectum is $4a$.

Substituting the value of $a$ from (i):

Length of latus rectum = $4 \times 2 = 8$

Therefore, for the parabola $x^2 = -8y$:

Coordinates of the focus are $(0, -2)$.

Equation of the directrix is $\mathbf{y = 2}$.

Length of the latus rectum is $8$.

Given:

The equation of the ellipse is $9x^2 + 25y^2 = 225$.

To Find:

Length of major and minor axes, eccentricity, coordinates of foci, and coordinates of vertices.

Solution:

The given equation of the ellipse is:

To convert this equation into the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, we divide the entire equation by 225:

Simplifying the fractions:

This is the standard form of an ellipse centered at the origin. Comparing equation (i) with the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, we have:

Taking the square root:

And

Taking the square root:

Since $a^2 > b^2$ (or $a > b$), the major axis is along the x-axis.

1. Major and Minor Axes:

Length of major axis = $2a$

Using the value of $a$ from (ii):

Length of major axis = $2 \times 5 = 10$

Length of minor axis = $2b$

Using the value of $b$ from (iii):

Length of minor axis = $2 \times 3 = 6$

2. Eccentricity ($e$):

For an ellipse with major axis along the x-axis, the eccentricity is given by $e = \sqrt{1 - \frac{b^2}{a^2}}$.

Substituting the values of $a^2$ and $b^2$:

3. Foci:

For an ellipse with major axis along the x-axis, the coordinates of the foci are $(\pm c, 0)$, where $c = ae$.

Using the values of $a$ from (ii) and $e$ from (iv):

The coordinates of the foci are $(\pm 4, 0)$.

4. Vertices:

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

Using the value of $a$ from (ii):

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

Summary of the results:

Length of major axis = $10$

Length of minor axis = $6$

Eccentricity ($e$) = $\mathbf{\frac{4}{5}}$ or $\mathbf{0.8}$

Coordinates of foci = $(\pm 4, 0)$

Coordinates of vertices = $(\pm 5, 0)$

Given:

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

Equation of one directrix is $x = \frac{36}{5}$.

To Find:

The equation of the ellipse.

Solution:

Since the foci are located at $(\pm 5, 0)$, the center of the ellipse is at the origin $(0,0)$, and the major axis lies along the x-axis.

For an ellipse centered at the origin with major axis on the x-axis, the coordinates of the foci are $(\pm c, 0)$, where $c$ is the distance from the center to the focus.

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

The equation of the directrix for an ellipse with major axis along the x-axis is $x = \pm \frac{a}{e}$, where $a$ is the semi-major axis and $e$ is the eccentricity.

Given one directrix is $x = \frac{36}{5}$. Therefore:

We also know the relationship between $a$, $c$, and $e$ for an ellipse is $c = ae$.

Substituting the value of $c=5$:

Now we have a system of two equations with two unknowns, $a$ and $e$.

From equation (ii), we can express $e$ in terms of $a$:

Substitute this expression for $e$ into equation (i):

Multiply both sides by 5:

Taking the square root (since $a$ must be positive):

Now we find $b^2$ using the relationship $c^2 = a^2 - b^2$ for an ellipse with major axis along the x-axis. Rearranging the formula to solve for $b^2$:

Substitute the values $a=6$ and $c=5$:

The standard equation of an ellipse centered at the origin with major axis along the x-axis is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.

Substitute the values of $a^2=36$ and $b^2=11$ into the standard equation:

The equation of the ellipse is $\mathbf{\frac{x^2}{36} + \frac{y^2}{11} = 1}$.

Given:

The equation of the hyperbola is $9x^2 - 16y^2 = 144$.

To Find:

Coordinates of the vertices, coordinates of the foci, and the eccentricity of the hyperbola.

Solution:

The given equation of the hyperbola is:

To convert this equation into the standard form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ or $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$, we divide the entire equation by 144:

Simplifying the fractions:

This is the standard form of a hyperbola centered at the origin with the transverse axis along the x-axis. Comparing equation (i) with the standard form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, we have:

Taking the square root (since $a$ must be positive):

And

Taking the square root (since $b$ must be positive):

1. Vertices:

For a hyperbola of the form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, the coordinates of the vertices are $(\pm a, 0)$.

Using the value of $a$ from (ii):

Vertices = $(\pm 4, 0)$

2. Foci:

For a hyperbola of the form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, the coordinates of the foci are $(\pm c, 0)$, where $c^2 = a^2 + b^2$.

Substitute the values of $a^2$ and $b^2$:

Taking the square root (since $c$ must be positive):

Using the value of $c$ from (iv):

Foci = $(\pm 5, 0)$

3. Eccentricity ($e$):

For a hyperbola, the eccentricity is given by $e = \frac{c}{a}$.

Using the values of $c$ from (iv) and $a$ from (ii):

Summary of the results:

Coordinates of the vertices are $(\pm 4, 0)$.

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

Eccentricity ($e$) is $\mathbf{\frac{5}{4}}$ or $\mathbf{1.25}$.

Given:

Coordinates of the vertices are $(0, \pm 6)$.

Eccentricity $e = \frac{5}{3}$.

To Find:

The equation of the hyperbola and the coordinates of its foci.

Solution:

Since the vertices are located at $(0, \pm 6)$, the hyperbola is centered at the origin $(0,0)$ and its transverse axis lies along the y-axis.

The standard form of a hyperbola with its transverse axis along the y-axis is $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$.

For this standard form, the coordinates of the vertices are $(0, \pm a)$.

Comparing the given vertices $(0, \pm 6)$ with $(0, \pm a)$, we get:

So, $a^2 = 6^2 = 36$.

The eccentricity of a hyperbola is given by $e = \frac{c}{a}$, where $c$ is the distance from the center to the focus.

We are given $e = \frac{5}{3}$ and we found $a = 6$. Substituting these values into the eccentricity formula:

Multiplying both sides by 6 to solve for $c$:

For a hyperbola, the relationship between $a$, $b$, and $c$ is $c^2 = a^2 + b^2$.

We need to find $b^2$ to write the equation of the hyperbola. Rearranging the formula:

Substitute the values of $c=10$ (from (i)) and $a=6$:

Now we can write the equation of the hyperbola using the standard form $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$ with $a^2 = 36$ and $b^2 = 64$:

For a hyperbola with transverse axis along the y-axis, the coordinates of the foci are $(0, \pm c)$.

Using the value of $c=10$ from (i):

The coordinates of the foci are $(0, \pm 10)$.

Summary of the results:

The equation of the hyperbola is $\mathbf{\frac{y^2}{36} - \frac{x^2}{64} = 1}$.

The coordinates of the foci are $(0, \pm 10)$.

Given:

The three points through which the circle passes are $(20, 3)$, $(19, 8)$, and $(2, -9)$.

To Find:

The equation of the circle, its center, and its radius.

Solution:

The general equation of a circle is given by $x^2 + y^2 + 2gx + 2fy + c = 0$, where $(-g, -f)$ is the center and $\sqrt{g^2 + f^2 - c}$ is the radius.

Since the circle passes through the given points, the coordinates of each point must satisfy the equation of the circle.

Substituting the point $(20, 3)$ into the general equation:

Substituting the point $(19, 8)$ into the general equation:

Substituting the point $(2, -9)$ into the general equation:

Now we have a system of three linear equations in $g$, $f$, and $c$.

Subtract equation (2) from equation (1):

Dividing by 2:

Subtract equation (3) from equation (2):

Dividing by 34:

Now we solve the system of equations (4) and (5).

From equation (5), $g = -10 - f$. Substitute this into equation (4):

Substitute the value of $f = -3$ into equation (5):

Now substitute the values of $g = -7$ and $f = -3$ into equation (3) to find $c$:

The equation of the circle is $x^2 + y^2 + 2gx + 2fy + c = 0$. Substituting the values of $g$, $f$, and $c$:

The center of the circle is $(-g, -f)$. Using $g = -7$ and $f = -3$:

The radius of the circle is $r = \sqrt{g^2 + f^2 - c}$. Using $g = -7$, $f = -3$, and $c = -111$:

The equation of the circle is $\mathbf{x^2 + y^2 - 14x - 6y - 111 = 0}$.

The coordinates of the center are $(7, 3)$.

The radius is $13$.

Given:

The equation of the parabola is $y^2 = 4ax$.

An equilateral triangle is inscribed in the parabola with one vertex at the vertex of the parabola.

To Find:

The length of the side of the equilateral triangle.

Solution:

The equation of the given parabola is $y^2 = 4ax$. The vertex of this parabola is at the origin, O $(0,0)$.

Let the equilateral triangle be OAB, where O is the vertex $(0,0)$. The other two vertices, A and B, lie on the parabola $y^2 = 4ax$.

Due to the symmetry of the parabola about the x-axis, if A is a point $(x_1, y_1)$ on the parabola, then the point $(x_1, -y_1)$ is also on the parabola. For the triangle OAB to be equilateral with O at the origin, the vertices A and B must be symmetric with respect to the x-axis.

Let the coordinates of vertex A be $(x_1, y_1)$ where $y_1 \neq 0$ (otherwise A would coincide with O). Since A lies on the parabola:

The coordinates of vertex B are $(x_1, -y_1)$. Since B also lies on the parabola:

This is the same condition as for A, confirming the symmetry.

The side lengths of the equilateral triangle OAB are OA, OB, and AB. These lengths must be equal.

Length of side OA = Distance between $(0,0)$ and $(x_1, y_1)$

$OA = \sqrt{(x_1 - 0)^2 + (y_1 - 0)^2} = \sqrt{x_1^2 + y_1^2}$

Length of side AB = Distance between $(x_1, y_1)$ and $(x_1, -y_1)$

$AB = \sqrt{(x_1 - x_1)^2 + (y_1 - (-y_1))^2} = \sqrt{0^2 + (2y_1)^2} = \sqrt{4y_1^2} = |2y_1|$

Since the triangle is equilateral, $OA^2 = AB^2$.

Now substitute the value of $y_1^2$ from equation (1) into equation (2):

Rearrange the equation:

Factor out $x_1$:

This gives two possible solutions for $x_1$:

$x_1 = 0$ or $x_1 = 12a$.

If $x_1 = 0$, then from equation (1), $y_1^2 = 4a(0) = 0$, which means $y_1 = 0$. This corresponds to the point $(0,0)$, the vertex O. Since A and B are distinct vertices of the triangle and not the vertex O, $x_1$ cannot be 0.

Therefore, the only valid solution for $x_1$ is $x_1 = 12a$.

Now find the value of $y_1^2$ using equation (1) with $x_1 = 12a$:

The length of the side of the equilateral triangle is $AB = |2y_1|$. We can calculate this using $y_1^2 = 48a^2$.

$|y_1| = \sqrt{48a^2} = \sqrt{16 \times 3 \times a^2} = \sqrt{16} \times \sqrt{3} \times \sqrt{a^2} = 4\sqrt{3}|a|$.

Side length = $AB = 2|y_1| = 2(4\sqrt{3}|a|) = 8\sqrt{3}|a|$.

Alternatively, using $OA = \sqrt{x_1^2 + y_1^2}$ with $x_1 = 12a$ and $y_1^2 = 48a^2$:

Since the length must be positive, we use the absolute value of $a$. In the standard representation $y^2=4ax$, $a$ is often taken as a positive parameter determining the focal length. If $a$ can be negative, the parabola opens left, and the length depends on $|a|$.

The length of the side of the equilateral triangle is $\mathbf{8\sqrt{3}|a|}$.

Given:

The ellipse passes through the point $(-3, 1)$.

The eccentricity is $e = \frac{\sqrt{2}}{5}$.

The major axis is along the x-axis, and the center is at the origin $(0,0)$.

To Find:

The equation of the ellipse.

Solution:

Since the center of the ellipse is at the origin and the major axis is along the x-axis, the standard equation of the ellipse is of the form:

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

The relationship between the eccentricity $e$, $a$, and $b$ for this type of ellipse is given by $e^2 = 1 - \frac{b^2}{a^2}$.

We are given the eccentricity $e = \frac{\sqrt{2}}{5}$. So, $e^2 = \left(\frac{\sqrt{2}}{5}\right)^2 = \frac{2}{25}$.

Substituting the value of $e^2$ into the relationship:

Rearranging the terms to express $b^2$ in terms of $a^2$:

The ellipse passes through the point $(-3, 1)$. This means the coordinates $x = -3$ and $y = 1$ must satisfy the equation of the ellipse:

Now, substitute the expression for $b^2$ from equation (1) into equation (2):

Combine the terms on the left side by finding a common denominator, which is $23a^2$:

Multiply both sides by $23a^2$:

Solve for $a^2$:

Now find $b^2$ using equation (1) and the value of $a^2$:

Substitute the values of $a^2 = \frac{232}{23}$ and $b^2 = \frac{232}{25}$ into the standard equation of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$:

Rewrite the terms:

Multiply the entire equation by 232 to clear the denominators:

The equation of the ellipse is $\mathbf{23x^2 + 25y^2 = 232}$.

Given:

Coordinates of the vertices are $(\pm 6, 0)$.

Equation of one directrix is $x = 4$.

To Find:

The equation of the hyperbola.

Solution:

Since the vertices are located at $(\pm 6, 0)$, the center of the hyperbola is at the origin $(0,0)$, and the transverse axis lies along the x-axis.

The standard form of a hyperbola centered at the origin with the transverse axis along the x-axis is:

where $a$ is the length of the semi-transverse axis.

For this standard form, the coordinates of the vertices are $(\pm a, 0)$.

Comparing the given vertices $(\pm 6, 0)$ with $(\pm a, 0)$, we get:

So, $a^2 = 6^2 = 36$.

The equation of the directrices for a hyperbola with the transverse axis along the x-axis is $x = \pm \frac{a}{e}$, where $e$ is the eccentricity.

Given one directrix is $x = 4$. Therefore:

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

Solving for the eccentricity $e$:

For a hyperbola, the relationship between $a$, $b$ (semi-conjugate axis length), and $e$ is $e^2 = 1 + \frac{b^2}{a^2}$.

Rearrange the formula to solve for $\frac{b^2}{a^2}$:

Substitute the value of $e = \frac{3}{2}$ from (ii):

Now, substitute the value of $a^2 = 36$:

Multiply both sides by 36 to solve for $b^2$:

Substitute the values of $a^2 = 36$ and $b^2 = 45$ into the standard equation of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$:

The equation of the hyperbola is $\mathbf{\frac{x^2}{36} - \frac{y^2}{45} = 1}$.

Given:

A circle in the first quadrant touching each coordinate axis at a distance of one unit from the origin.

To Find:

The equation of the circle.

Solution:

Since the circle touches the positive x-axis at a distance of one unit from the origin, the point of tangency on the x-axis is $(1, 0)$.

The radius of the circle is the perpendicular distance from the center to the tangent line (x-axis). If the center is $(h, k)$, the radius $r$ is $|k|$. The line joining the center to the point of tangency $(1, 0)$ must be perpendicular to the x-axis. Thus, the x-coordinate of the center must be 1. So the center is $(1, k)$ and the radius $r = |k|$. Since the circle is in the first quadrant, $k > 0$, so $r=k$. The center is $(1, r)$.

Since the circle touches the positive y-axis at a distance of one unit from the origin, the point of tangency on the y-axis is $(0, 1)$.

The radius of the circle is the perpendicular distance from the center to the tangent line (y-axis). If the center is $(h, k)$, the radius $r$ is $|h|$. The line joining the center to the point of tangency $(0, 1)$ must be perpendicular to the y-axis. Thus, the y-coordinate of the center must be 1. So the center is $(h, 1)$ and the radius $r = |h|$. Since the circle is in the first quadrant, $h > 0$, so $r=h$. The center is $(r, 1)$.

Comparing the coordinates of the center from both conditions, we have $h=r$ and $k=r$. Substituting the values we found, the center is $(1, 1)$, and the radius is $r=1$.

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

Substitute the center $(1, 1)$ and radius $r=1$ into equation (i):

Expand the terms:

Rearrange the terms to get the standard form of the equation:

This equation matches option (A).

The correct answer is (A) $x^2 + y^2 – 2x – 2y + 1= 0$.

Given:

Center of the circle is $(1, -2)$.

The circle passes through the point of intersection of the lines $3x + y = 14$ and $2x + 5y = 18$.

To Find:

The equation of the circle.

Solution:

Let the center of the circle be $(h, k)$. We are given that $(h, k) = (1, -2)$.

First, we need to find the point of intersection of the given lines:

From equation (i), we can express $y$ in terms of $x$:

Substitute equation (iii) into equation (ii):

Now substitute the value of $x=4$ into equation (iii):

So, the point of intersection of the lines is $(4, 2)$.

The circle passes through this point $(4, 2)$. The radius $r$ of the circle is the distance between the center $(1, -2)$ and the point $(4, 2)$.

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

The equation of a circle with center $(h, k)$ and radius $r$ is $(x - h)^2 + (y - k)^2 = r^2$.

Substitute the center $(1, -2)$ and radius $r=5$ into the equation:

Expand the terms:

Rearrange the terms to set the equation to zero:

This equation matches option (A).

The correct answer is (A) $x^2 + y^2 – 2x + 4y – 20 = 0$.

Given:

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

The triangle is formed by the vertex of the parabola and the ends of its latus rectum.

To Find:

The area of the triangle.

Solution:

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

This equation is of the form $x^2 = 4ay$, where $4a = 12$.

From $4a = 12$, we get $a = 3$.

The vertex of the parabola $x^2 = 4ay$ is at the origin $(0, 0)$.

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

The latus rectum is a line segment perpendicular to the axis of the parabola (y-axis) passing through the focus.

The equation of the latus rectum is $y = a$, which is $y = 3$.

To find the endpoints of the latus rectum, substitute $y = 3$ into the parabola equation $x^2 = 12y$:

So, the endpoints of the latus rectum are $(-6, 3)$ and $(6, 3)$.

The triangle is formed by the vertices V$(0, 0)$, A$(-6, 3)$, and B$(6, 3)$.

We can find the area of this triangle using the base and height.

The base of the triangle can be the distance between the points A and B.

The base AB lies on the line $y=3$. The height of the triangle is the perpendicular distance from the vertex V$(0, 0)$ to the line $y=3$.

The area of the triangle VAB is given by:

Alternatively, using the determinant formula for the area of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$:

Using V$(0, 0)$, A$(-6, 3)$, B$(6, 3)$: $(x_1, y_1) = (0, 0)$, $(x_2, y_2) = (-6, 3)$, $(x_3, y_3) = (6, 3)$.

Both methods give the same result.

The correct answer is (C) 18 sq. units.

Given:

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

The lines join the vertex of the parabola to the points on it which have abscissa 24.

To Find:

The equations of the lines.

Solution:

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

This equation is of the form $y^2 = 4ax$, where $4a = 6$. Thus, $a = \frac{6}{4} = \frac{3}{2}$.

The vertex of the parabola $y^2 = 4ax$ is at the origin $(0, 0)$.

We are given that the points on the parabola have an abscissa (x-coordinate) of 24. Let the points be $(24, y)$.

Substitute $x = 24$ into the equation of the parabola $y^2 = 6x$ to find the corresponding y-coordinates:

Taking the square root on both sides:

So, the points on the parabola with abscissa 24 are $P_1(24, 12)$ and $P_2(24, -12)$.

We need to find the equations of the lines joining the vertex V$(0, 0)$ to these points.

Equation of the line joining V$(0, 0)$ and $P_1(24, 12)$:

The slope of the line passing through $(x_1, y_1)$ and $(x_2, y_2)$ is $m = \frac{y_2 - y_1}{x_2 - x_1}$.

The equation of the line passing through $(0, 0)$ with slope $m_1$ is $y - 0 = m_1(x - 0)$, i.e., $y = m_1x$.

Multiply by 2:

Rearrange the terms:

Equation of the line joining V$(0, 0)$ and $P_2(24, -12)$:

The slope of the line passing through $(0, 0)$ and $(24, -12)$ is:

The equation of the line passing through $(0, 0)$ with slope $m_2$ is $y - 0 = m_2(x - 0)$, i.e., $y = m_2x$.

Multiply by 2:

Rearrange the terms:

The equations of the two lines are $x - 2y = 0$ and $x + 2y = 0$.

These two equations can be combined and written in the form $x \pm 2y = 0$.

Let's check the given options:

(A) $y \pm 2x = 0 \implies y = 2x$ or $y = -2x$. (Slopes $\pm 2$)

(B) $2y \pm x = 0 \implies 2y = -x$ or $2y = x \implies y = -\frac{1}{2}x$ or $y = \frac{1}{2}x$. (Slopes $\pm \frac{1}{2}$)

(C) $x \pm 2y = 0 \implies x = -2y$ or $x = 2y \implies y = -\frac{1}{2}x$ or $y = \frac{1}{2}x$. (Slopes $\pm \frac{1}{2}$)

(D) $2x \pm y = 0 \implies y = -2x$ or $y = 2x$. (Slopes $\mp 2$)

Both options (B) and (C) mathematically represent the same pair of lines $x+2y=0$ and $x-2y=0$. However, option (C) directly matches the form $x \pm 2y = 0$ derived from equations (i) and (ii).

The correct answer is (C) $x \pm 2y = 0$.

Given:

The center of the ellipse is at the origin $(0, 0)$.

The x-axis is the major axis.

The ellipse passes through the points $(-3, 1)$ and $(2, -2)$.

To Find:

The equation of the ellipse.

Solution:

Since the center of the ellipse is at the origin $(0, 0)$ and the x-axis is the major axis, the standard equation of the ellipse is:

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

The ellipse also passes through the point $(2, -2)$. Substituting $x = 2$ and $y = -2$ into the equation:

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

Let $u = \frac{1}{a^2}$ and $v = \frac{1}{b^2}$. The system becomes:

From equation (iii), express $v$ in terms of $u$:

Substitute equation (v) into equation (iv):

Now substitute the value of $u = \frac{3}{32}$ into equation (v):

We have $u = \frac{1}{a^2} = \frac{3}{32}$, so $a^2 = \frac{32}{3}$.

We have $v = \frac{1}{b^2} = \frac{5}{32}$, so $b^2 = \frac{32}{5}$.

Check that $a^2 > b^2$: $\frac{32}{3} > \frac{32}{5}$ since $3 < 5$. This confirms that the x-axis is indeed the major axis.

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

Multiply both sides of the equation by 32:

This equation matches option (B).

The correct answer is (B) $3x^2 + 5y^2 = 32$.

Given:

The center of the hyperbola is at the origin $(0, 0)$.

The transverse axis is along the x-axis.

The length of the transverse axis is 7 units.

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

To Find:

The equation of the hyperbola.

Solution:

Since the center of the hyperbola is at the origin $(0, 0)$ and the transverse axis is along the x-axis, the standard equation of the hyperbola is:

Here, $a$ is the semi-transverse axis length and $b$ is the semi-conjugate axis length.

The length of the transverse axis is given as 7. The length of the transverse axis for this standard equation is $2a$.

Squaring $a$ gives $a^2$:

Substitute the value of $a^2$ into equation (i):

The hyperbola passes through the point $(5, -2)$. Substitute $x = 5$ and $y = -2$ into equation (ii):

Rearrange the terms to solve for $b^2$:

Cross-multiply to find $b^2$:

Now substitute the values of $a^2 = \frac{49}{4}$ and $b^2 = \frac{196}{51}$ back into the standard equation $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$:

Simplify the expression:

Let's compare this equation with the given options:

(A) $\frac{4}{49} x^2 - \frac{196}{51} y^2 = 1$. This is $\frac{x^2}{49/4} - \frac{y^2}{51/196} = 1$. This does not match (iii).

(B) $\frac{49}{4} x^2 - \frac{51}{196} y^2 = 1$. This is $\frac{x^2}{4/49} - \frac{y^2}{196/51} = 1$. This does not match (iii).

(C) $\frac{4}{49} x^2 - \frac{51}{196} y^2 = 1$. This matches equation (iii).

(D) none of these.

The correct answer is (C) $\frac{4}{49} x^2 - \frac{51}{196} y^2 = 1$.

Statement:

The given statement is correct.

Justification:

The coordinates of any point on the circle are given by the parametric equations:

From equation (i), we can write:

From equation (ii), we can write:

Square both equations (iii) and (iv):

Add equations (v) and (vi):

Using the trigonometric identity $\cos^2 \theta + \sin^2 \theta = 1$:

This is the standard equation of a circle with center $(2, -1)$ and radius $\sqrt{16} = 4$.

The derived equation (vii) is identical to the equation given in the statement.

Therefore, the statement that the equation of the circle on which the coordinates of any point are $(2 + 4 \cos \theta , –1 + 4 \sin \theta)$ is $(x – 2)^2 + (y + 1)^2 = 16$ is correct.

Statement:

The given statement is correct.

Justification:

Let the given length of the bar be $L$. Let the two fixed straight lines at right angles be the x-axis and the y-axis. Let the ends of the bar touching the axes be $A$ and $B$.

Let $A$ be on the x-axis and $B$ be on the y-axis. So, the coordinates of $A$ are $(x_A, 0)$ and the coordinates of $B$ are $(0, y_B)$.

The length of the bar $AB$ is constant and equal to $L$. Using the distance formula, we have:

Now, consider any point $P(x, y)$ on the bar $AB$. Let this point $P$ divide the bar such that the distance from $A$ to $P$ is $b$ and the distance from $P$ to $B$ is $a$. Thus, the bar's length $L = AP + PB = b + a$. Note that $a$ and $b$ are positive constants for a specific point $P$ on the bar.

Using the section formula, the coordinates of a point $P(x, y)$ dividing the line segment joining $A(x_A, 0)$ and $B(0, y_B)$ in the ratio $PB : AP = a : b$ are given by:

From these equations, we can express $x_A$ and $y_B$ in terms of $x, y, a, b, L$:

Substitute equations (ii) and (iii) into equation (i):

Since $L \neq 0$, we can divide the entire equation by $L^2$:

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

The semi-axes of this ellipse are $b$ and $a$. Since $a$ and $b$ are positive constants for any fixed point $P$ on the bar (unless $P$ is one of the endpoints), the locus of $P$ is an ellipse.

If the point $P$ is the midpoint of the bar, then $a=b=L/2$. In this case, the equation becomes $\frac{x^2}{(L/2)^2} + \frac{y^2}{(L/2)^2} = 1$, which simplifies to $x^2 + y^2 = (L/2)^2$. This is the equation of a circle with radius $L/2$, which is a special case of an ellipse.

Therefore, any point on the bar (including the midpoint) describes an ellipse as its extremities move along the two fixed perpendicular lines.

Given:

Centre of the circle is $(2, 2)$.

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

To Find:

The equation of the circle.

Solution:

Let the centre of the circle be $(h, k) = (2, 2)$.

Let the point the circle passes through be $(x_1, y_1) = (4, 5)$.

The radius $r$ of the circle is the distance between its centre $(h, k)$ and any point $(x_1, y_1)$ on the circle.

Using the distance formula $r = \sqrt{(x_1 - h)^2 + (y_1 - k)^2}$:

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

This is the required equation of the circle.

The equation of the circle which passes through the point (4, 5) and has its centre at (2, 2) is $(\text{x} - 2)^\textbf{2} + (\text{y} - 2)^\textbf{2} = 13$.

Given:

Radius of the circle $r = 3$ units.

The centre of the circle lies on the line $y = x - 1$.

The circle passes through the point $(7, 3)$.

To Find:

The equation of the circle.

Solution:

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

Since the centre $(h, k)$ lies on the line $y = x - 1$, its coordinates satisfy the equation of the line:

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

We are given that the radius $r = 3$. So, the equation is $(x - h)^2 + (y - k)^2 = 3^2 = 9$.

The circle passes through the point $(7, 3)$. This means the coordinates of this point must satisfy the equation of the circle. Substitute $x = 7$ and $y = 3$ into equation (ii):

Now, substitute the relation $k = h - 1$ from equation (i) into equation (iii):

Expand the squared terms:

Combine like terms:

Move the constant term to the left side:

Divide the entire equation by 2:

Factor the quadratic equation:

This gives two possible values for $h$:

Now find the corresponding values of $k$ using $k = h - 1$:

If $h = 4$, then $k = 4 - 1 = 3$. The centre is $(4, 3)$.

If $h = 7$, then $k = 7 - 1 = 6$. The centre is $(7, 6)$.

We have two possible centres, each giving a valid circle with radius 3 passing through $(7, 3)$.

Case 1: Centre $(4, 3)$ and radius $r = 3$.

The equation of the circle is $(x - 4)^2 + (y - 3)^2 = 3^2 = 9$.

Case 2: Centre $(7, 6)$ and radius $r = 3$.

The equation of the circle is $(x - 7)^2 + (y - 6)^2 = 3^2 = 9$.

Both equations (iv) and (v) are valid answers.

A circle has radius 3 units and its centre lies on the line y = x – 1. If it passes through the point (7, 3), its equation is $(\text{x} - 4)^\textbf{2} + (\text{y} - 3)^\textbf{2} = 9$ or $(\text{x} - 7)^\textbf{2} + (\text{y} - 6)^\textbf{2} = 9$.

Given:

Ellipse with centre at origin and axis along x-axis (Major axis is x-axis).

Length of the latus rectum is 10.

Distance between foci = Length of minor axis.

To Find:

The equation of the ellipse.

Solution:

Since the centre of the ellipse is at the origin $(0, 0)$ and the major axis is along the x-axis, the standard equation of the ellipse is:

The length of the latus rectum of this ellipse is $\frac{2b^2}{a}$. We are given that the length of the latus rectum is 10.

This simplifies to:

The distance between the foci of the ellipse is $2ae$, where $e$ is the eccentricity. The foci are at $(\pm ae, 0)$.

The length of the minor axis is $2b$.

We are given that the distance between foci is equal to the length of the minor axis:

This simplifies to:

Square both sides of equation (ii):

For an ellipse with major axis along the x-axis, the relation between $a, b,$ and $e$ is $b^2 = a^2(1 - e^2)$.

Substitute $b^2 = a^2 e^2$ from equation (iii) into this relation:

Since $a \neq 0$ (as the latus rectum has a length), we can divide both sides by $a^2$:

Rearrange the terms to solve for $e^2$:

Now substitute the value of $e^2$ back into equation (iii), $b^2 = a^2 e^2$:

Now we have two equations relating $a$ and $b^2$: equation (i) $b^2 = 5a$ and equation (iv) $b^2 = \frac{a^2}{2}$.

Equate the expressions for $b^2$:

Multiply both sides by 2:

Rearrange the terms to form a quadratic equation in $a$:

Factor out $a$:

This gives two possible values for $a$: $a = 0$ or $a = 10$. Since $a$ is the length of the semi-major axis, $a$ must be positive. So, $a = 10$.

Now substitute the value of $a=10$ into equation (i) to find $b^2$:

So, $a^2 = 10^2 = 100$ and $b^2 = 50$. We confirm that $a^2 > b^2$ ($100 > 50$), which is consistent with the major axis being along the x-axis.

Substitute the values of $a^2$ and $b^2$ into the standard equation of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$:

If the latus rectum of an ellipse with axis along x-axis and centre at origin is 10, distance between foci = length of minor axis, then the equation of the ellipse is $\frac{\text{x}^2}{100} + \frac{\text{y}^2}{50} = 1$.

Given:

Focus $F = (2, 3)$.

Directrix is the line $x - 4y + 3 = 0$.

To Find:

The equation of the parabola.

Solution:

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

By the definition of a parabola, every point on the parabola is equidistant from the focus and the directrix.

Distance from point $P(x, y)$ to the Focus $F(2, 3)$ is $PF$. Using the distance formula:

The perpendicular distance from point $P(x, y)$ to the Directrix $x - 4y + 3 = 0$ is $PD$. Using the formula for the perpendicular distance from a point $(x_0, y_0)$ to a line $Ax + By + C = 0$, which is $\frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$:

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

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

Multiply both sides by 17:

Expand the squared terms on both sides:

Distribute 17 on the left side:

Move all terms to the left side to set the equation to zero:

Combine the like terms:

The equation of the parabola whose focus is the point (2, 3) and directrix is the line x – 4y + 3 = 0 is $16x^2 + y^2 + 8xy - 74x - 78y + 212 = 0$.

Given:

The equation of the hyperbola is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.

The hyperbola passes through the points $(3, 0)$ and $(3 \sqrt{2} , 2)$.

To Find:

The eccentricity of the hyperbola.

Solution:

The standard equation of the hyperbola with the transverse axis along the x-axis and centre at the origin is given by:

The hyperbola passes through the point $(3, 0)$. Substitute $x=3$ and $y=0$ into equation (i):

The hyperbola passes through the point $(3 \sqrt{2}, 2)$. Substitute $x=3 \sqrt{2}$, $y=2$, and $a^2=9$ from equation (ii) into equation (i):

Rearrange the terms to solve for $b^2$:

The eccentricity $e$ of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is given by the formula:

Substitute the values of $a^2 = 9$ (from equation ii) and $b^2 = 4$ (from equation iii) into equation (iv):

The eccentricity of the hyperbola $\frac{x^2}{a^2} \;-\; \frac{y^2}{b^2} = 1$ which passes through the points (3, 0) and $(3 \sqrt{2} , 2)$ is $\frac{\sqrt{13}}{3}$.

Given:

The circle touches both the x-axis and the y-axis in the first quadrant.

The radius of the circle is $a$.

To Find:

The equation of the circle.

Solution:

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

We are given that the radius is $a$, so $r = a$.

Since the circle touches the x-axis, the distance from the center $(h, k)$ to the x-axis is equal to the radius. The distance is $|k|$. Thus, $|k| = r = a$.

Since the circle touches the y-axis, the distance from the center $(h, k)$ to the y-axis is equal to the radius. The distance is $|h|$. Thus, $|h| = r = a$.

The circle lies in the first quadrant. This implies that the center $(h, k)$ is in the first quadrant, so both $h$ and $k$ are positive.

Therefore, from $|h| = a$ and $h > 0$, we have $h = a$.

Similarly, from $|k| = a$ and $k > 0$, we have $k = a$.

So, the center of the circle is $(a, a)$ and the radius is $a$.

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

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

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

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

This is the required equation of the circle.

We can also expand this equation:

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

$x^2 + y^2 - 2ax - 2ay + a^2 + a^2 - a^2 = 0$

$x^2 + y^2 - 2ax - 2ay + a^2 = 0$

Both forms $(x-a)^2 + (y-a)^2 = a^2$ and $x^2 + y^2 - 2ax - 2ay + a^2 = 0$ represent the equation of the circle.

Given:

The coordinates of a point $(x, y)$ are given by the parametric equations:

$x = \frac{2at}{1 + t^2}$

$y = \frac{a(1 − t^2)}{1 + t^2}$

where $t$ is a real number such that $-1 \leq t \leq 1$, and $a$ is a given real number.

To Show:

The point $(x, y)$ lies on a circle for all real values of $t$ such that $-1 \leq t \leq 1$.

Solution:

To show that the point $(x, y)$ lies on a circle, we need to find an equation relating $x$ and $y$ that does not involve the parameter $t$ and represents the equation of a circle.

Let's calculate the square of $x$:

$x^2 = \left( \frac{2at}{1 + t^2} \right)^2 = \frac{(2at)^2}{(1 + t^2)^2} = \frac{4a^2t^2}{(1 + t^2)^2}$

Now, let's calculate the square of $y$:

$y^2 = \left( \frac{a(1 - t^2)}{1 + t^2} \right)^2 = \frac{a^2(1 - t^2)^2}{(1 + t^2)^2}$

Expand the term $(1 - t^2)^2$ in the numerator of $y^2$:

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

Now, let's add $x^2$ and $y^2$:

$x^2 + y^2 = \frac{4a^2t^2}{(1 + t^2)^2} + \frac{a^2(1 - 2t^2 + t^4)}{(1 + t^2)^2}$

Since the denominators are the same, we can add the numerators:

$x^2 + y^2 = \frac{4a^2t^2 + a^2(1 - 2t^2 + t^4)}{(1 + t^2)^2}$

Distribute $a^2$ in the numerator:

$x^2 + y^2 = \frac{4a^2t^2 + a^2 - 2a^2t^2 + a^2t^4}{(1 + t^2)^2}$

Combine the terms in the numerator:

$x^2 + y^2 = \frac{a^2t^4 + (4a^2 - 2a^2)t^2 + a^2}{(1 + t^2)^2}$

$x^2 + y^2 = \frac{a^2t^4 + 2a^2t^2 + a^2}{(1 + t^2)^2}$

Factor out $a^2$ from the numerator:

$x^2 + y^2 = \frac{a^2(t^4 + 2t^2 + 1)}{(1 + t^2)^2}$

Recognize that the expression inside the parenthesis in the numerator, $t^4 + 2t^2 + 1$, is a perfect square, which can be written as $(t^2 + 1)^2$ or $(1 + t^2)^2$.

$x^2 + y^2 = \frac{a^2(1 + t^2)^2}{(1 + t^2)^2}$

Since $t$ is a real number, $t^2 \geq 0$, which implies $1 + t^2 \geq 1$. Thus, $1 + t^2$ is never zero, and neither is $(1 + t^2)^2$. We can therefore cancel the term $(1 + t^2)^2$ from the numerator and the denominator.

$x^2 + y^2 = a^2$

This equation is in the form $x^2 + y^2 = r^2$, which is the standard equation of a circle centered at the origin $(0, 0)$ with radius $r = \sqrt{a^2} = |a|$.

The relation $x^2 + y^2 = a^2$ is independent of the parameter $t$. This shows that for any real value of $t$, the point $(x, y)$ lies on the circle defined by this equation.

The constraint $-1 \leq t \leq 1$ means that as $t$ varies within this range, the point $(x, y)$ traces a specific part (an arc) of the circle $x^2 + y^2 = a^2$. However, for every valid value of $t$, the point $(x, y)$ generated by the parametric equations satisfies the equation of the circle.

Therefore, the point $(x, y)$ lies on a circle for all real values of $t$ such that $-1 \leq t \leq 1$.

Given:

A circle passes through the points $(0, 0)$, $(a, 0)$, and $(0, b)$.

To Find:

The coordinates of the center of the circle.

Solution:

Let the three points be A$(0, 0)$, B$(a, 0)$, and C$(0, b)$.

The center of the circle passing through three non-collinear points is the intersection point of the perpendicular bisectors of the chords formed by these points.

Consider the chord AB connecting the points $(0, 0)$ and $(a, 0)$. This chord lies on the x-axis.

The midpoint of the chord AB is given by:

$M = \left(\frac{0 + a}{2}, \frac{0 + 0}{2}\right) = \left(\frac{a}{2}, 0\right)$

Since the chord AB lies on the x-axis (a horizontal line), its perpendicular bisector will be a vertical line passing through the midpoint $M(\frac{a}{2}, 0)$.

The equation of the perpendicular bisector of chord AB is:

$x = \frac{a}{2}$

Consider the chord AC connecting the points $(0, 0)$ and $(0, b)$. This chord lies on the y-axis.

The midpoint of the chord AC is given by:

$N = \left(\frac{0 + 0}{2}, \frac{0 + b}{2}\right) = \left(0, \frac{b}{2}\right)$

Since the chord AC lies on the y-axis (a vertical line), its perpendicular bisector will be a horizontal line passing through the midpoint $N(0, \frac{b}{2})$.

The equation of the perpendicular bisector of chord AC is:

$y = \frac{b}{2}$

The center of the circle is the point of intersection of the perpendicular bisectors of the chords AB and AC.

The intersection of the lines $x = \frac{a}{2}$ and $y = \frac{b}{2}$ is the point $\left(\frac{a}{2}, \frac{b}{2}\right)$.

Therefore, the coordinates of the center of the circle are $\left(\frac{a}{2}, \frac{b}{2}\right)$.

Alternate Solution:

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

Since the circle passes through the point $(0, 0)$, substitute $x = 0$ and $y = 0$ into the equation:

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

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

$c = 0$

So, the equation of the circle becomes $x^2 + y^2 + 2gx + 2fy = 0$.

Since the circle passes through the point $(a, 0)$, substitute $x = a$ and $y = 0$ into the equation:

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

$a^2 + 2ga = 0$

$a(a + 2g) = 0$

Since the points are distinct, $a \neq 0$. Therefore,

$a + 2g = 0$

Since the circle passes through the point $(0, b)$, substitute $x = 0$ and $y = b$ into the equation:

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

$b^2 + 2fb = 0$

$b(b + 2f) = 0$

Since the points are distinct, $b \neq 0$. Therefore,

$b + 2f = 0$

The coordinates of the center of the circle with the equation $x^2 + y^2 + 2gx + 2fy + c = 0$ are $(-g, -f)$.

From equation (i), $2g = -a$, which means $g = -\frac{a}{2}$.

From equation (ii), $2f = -b$, which means $f = -\frac{b}{2}$.

The coordinates of the center are $(-g, -f) = \left(-\left(-\frac{a}{2}\right), -\left(-\frac{b}{2}\right)\right) = \left(\frac{a}{2}, \frac{b}{2}\right)$.

Thus, the coordinates of the center of the circle are $\left(\frac{a}{2}, \frac{b}{2}\right)$.

Given:

The center of the circle is $(1, 2)$.

The circle touches the x-axis.

To Find:

The equation of the circle.

Solution:

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

We are given that the center of the circle is $(1, 2)$. So, $h = 1$ and $k = 2$.

Since the circle touches the x-axis, the distance from the center $(h, k)$ to the x-axis is equal to the radius $r$.

The distance from a point $(h, k)$ to the x-axis is given by $|k|$.

Therefore, the radius $r$ of the circle is $|k| = |2| = 2$.

So, the radius is $r = 2$.

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

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

Substitute the values of $h = 1$, $k = 2$, and $r = 2$ into the standard equation:

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

$(x - 1)^2 + (y - 2)^2 = 4$

This is the equation of the circle.

We can also expand this equation:

$(x^2 - 2(x)(1) + 1^2) + (y^2 - 2(y)(2) + 2^2) = 4$

$(x^2 - 2x + 1) + (y^2 - 4y + 4) = 4$

$x^2 - 2x + 1 + y^2 - 4y + 4 = 4$

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

Subtract 4 from both sides:

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

$x^2 + y^2 - 2x - 4y + 1 = 0$

Both forms $(x - 1)^2 + (y - 2)^2 = 4$ and $x^2 + y^2 - 2x - 4y + 1 = 0$ are valid equations for the circle.

Given:

The equations of two tangent lines to a circle are:

Line 1: $3x - 4y + 4 = 0$

Line 2: $6x - 8y - 7 = 0$

To Find:

The radius of the circle.

Solution:

The given lines are $3x - 4y + 4 = 0$ and $6x - 8y - 7 = 0$.

We observe that the coefficients of $x$ and $y$ in the second equation are double those in the first equation ($6 = 2 \times 3$ and $-8 = 2 \times -4$). This indicates that the lines are parallel.

To find the distance between these parallel lines, we first rewrite the second equation by dividing by 2 to make the coefficients of $x$ and $y$ the same as in the first equation:

$\frac{6x - 8y - 7}{2} = \frac{0}{2}$

$3x - 4y - \frac{7}{2} = 0$

Now we have the two parallel lines in the form $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$:

$3x - 4y + 4 = 0$ (Here $A = 3$, $B = -4$, $C_1 = 4$)

$3x - 4y - \frac{7}{2} = 0$ (Here $A = 3$, $B = -4$, $C_2 = -\frac{7}{2}$)

The distance between two parallel lines $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$ is given by the formula:

$d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$

Substitute the values of $A$, $B$, $C_1$, and $C_2$ into the formula:

$d = \frac{|4 - (-\frac{7}{2})|}{\sqrt{3^2 + (-4)^2}}$

$d = \frac{|4 + \frac{7}{2}|}{\sqrt{9 + 16}}$

$d = \frac{|\frac{8}{2} + \frac{7}{2}|}{\sqrt{25}}$

$d = \frac{|\frac{15}{2}|}{5}$

$d = \frac{\frac{15}{2}}{5}$

$d = \frac{15}{2} \times \frac{1}{5}$

$d = \frac{\cancel{15}^{3}}{2 \times \cancel{5}_{1}}$

$d = \frac{3}{2}$

The distance between the two parallel tangent lines of a circle is equal to the diameter of the circle.

Diameter $= d = \frac{3}{2}$

The radius of the circle ($r$) is half of the diameter.

$r = \frac{\text{Diameter}}{2}$

$r = \frac{3/2}{2}$

$r = \frac{3}{2} \times \frac{1}{2}$

$r = \frac{3}{4}$

The radius of the circle is $\frac{3}{4}$.

Given:

The circle touches both the x-axis and the y-axis.

The circle lies in the third quadrant.

The circle touches the line $3x - 4y + 8 = 0$.

To Find:

The equation of the circle.

Solution:

Since the circle touches both the x-axis and the y-axis and lies in the third quadrant, its center must be at a distance equal to its radius from both axes. Let the radius of the circle be $a$. Since the circle is in the third quadrant, both the x and y coordinates of the center must be negative and equal to $-a$.

So, the center of the circle is $(-a, -a)$, and the radius is $a$, where $a > 0$ (since the radius is a length).

The circle also touches the line $3x - 4y + 8 = 0$. The distance from the center of the circle to this tangent line must be equal to the radius $a$.

The formula for the perpendicular distance from a point $(x_0, y_0)$ to a line $Ax + By + C = 0$ is:

$d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$

In this case, the point is the center $(-a, -a)$, and the line is $3x - 4y + 8 = 0$. So, $(x_0, y_0) = (-a, -a)$, $A = 3$, $B = -4$, and $C = 8$. The distance $d$ is equal to the radius $a$.

Substitute these values into the distance formula:

$a = \frac{|3(-a) - 4(-a) + 8|}{\sqrt{3^2 + (-4)^2}}$

$a = \frac{|-3a + 4a + 8|}{\sqrt{9 + 16}}$

$a = \frac{|a + 8|}{\sqrt{25}}$

$a = \frac{|a + 8|}{5}$

Since the circle is in the third quadrant, the radius $a$ must be positive ($a > 0$). This implies that $a + 8$ is also positive ($a + 8 > 8 > 0$). Therefore, $|a + 8| = a + 8$.

So, the equation becomes:

$a = \frac{a + 8}{5}$

Multiply both sides by 5:

$5a = a + 8$

Subtract $a$ from both sides:

$5a - a = 8$

$4a = 8$

Divide by 4:

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

$a = 2$

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

The center of the circle is $(-a, -a) = (-2, -2)$.

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

Substitute the center $(h, k) = (-2, -2)$ and radius $r = 2$ into the equation:

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

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

This is the equation of the circle.

We can expand this equation if needed:

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

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

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

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

Both $(x + 2)^2 + (y + 2)^2 = 4$ and $x^2 + y^2 + 4x + 4y + 4 = 0$ are valid equations for the circle.

Given:

The equation of the circle is $x^2 + y^2 – 4x – 6y + 11 = 0$.

One end of a diameter of the circle is the point $(3, 4)$.

To Find:

The coordinates of the other end of the diameter.

Solution:

The general equation of a circle is given by $x^2 + y^2 + 2gx + 2fy + c = 0$. The coordinates of the center of the circle are $(-g, -f)$.

Comparing the given equation $x^2 + y^2 – 4x – 6y + 11 = 0$ with the general equation, we have:

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

$2f = -6 \implies f = -3$

$c = 11$

The center of the circle is $(-g, -f) = (-(-2), -(-3)) = (2, 3)$.

Let the center of the circle be $C(2, 3)$.

Let the given end of the diameter be $A(3, 4)$.

Let the other end of the diameter be $B(x_2, y_2)$.

The center of the circle is the midpoint of any diameter.

Using the midpoint formula, the coordinates of the center $C(h, k)$ are given by $h = \frac{x_1 + x_2}{2}$ and $k = \frac{y_1 + y_2}{2}$, where $(x_1, y_1)$ and $(x_2, y_2)$ are the coordinates of the endpoints of the diameter.

Here, $(h, k) = (2, 3)$ and $(x_1, y_1) = (3, 4)$. We need to find $(x_2, y_2)$.

For the x-coordinate of the center:

$2 = \frac{3 + x_2}{2}$

Multiply both sides by 2:

$2 \times 2 = 3 + x_2$

$4 = 3 + x_2$

Subtract 3 from both sides:

$x_2 = 4 - 3$

$x_2 = 1$

For the y-coordinate of the center:

$3 = \frac{4 + y_2}{2}$

Multiply both sides by 2:

$3 \times 2 = 4 + y_2$

$6 = 4 + y_2$

Subtract 4 from both sides:

$y_2 = 6 - 4$

$y_2 = 2$

So, the coordinates of the other end of the diameter are $(1, 2)$.

Given:

The center of the circle is $(1, -2)$.

The circle passes through the intersection point of the lines $3x + y = 14$ and $2x + 5y = 18$.

To Find:

The equation of the circle.

Solution:

Let the center of the circle be $C(h, k) = (1, -2)$.

The circle passes through the intersection point of the lines $3x + y = 14$ and $2x + 5y = 18$. Let's find this intersection point.

We have the system of linear equations:

$3x + y = 14 \quad \text{(1)}$

$2x + 5y = 18 \quad \text{(2)}$

From equation (1), we can express $y$ in terms of $x$:

$y = 14 - 3x \quad \text{(3)}$

Substitute the expression for $y$ from equation (3) into equation (2):

$2x + 5(14 - 3x) = 18$

$2x + 70 - 15x = 18$

Combine like terms:

$-13x + 70 = 18$

Subtract 70 from both sides:

$-13x = 18 - 70$

$-13x = -52$

Divide by -13:

$x = \frac{-52}{-13} = 4$

Now substitute the value of $x$ back into equation (3) to find $y$:

$y = 14 - 3(4)$

$y = 14 - 12$

$y = 2$

So, the intersection point of the two lines is $P(4, 2)$. This point lies on the circle.

The radius of the circle ($r$) is the distance between the center $C(1, -2)$ and the point $P(4, 2)$.

Using the distance formula between two points $(x_1, y_1)$ and $(x_2, y_2)$ which is $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$:

$r = \sqrt{(4 - 1)^2 + (2 - (-2))^2}$

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

$r = \sqrt{3^2 + 4^2}$

$r = \sqrt{9 + 16}$

$r = \sqrt{25}$

$r = 5$

The radius of the circle is 5.

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

Substitute the center $(h, k) = (1, -2)$ and radius $r = 5$ into the equation:

$(x - 1)^2 + (y - (-2))^2 = 5^2$

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

This is the equation of the circle.

We can expand this equation if needed:

$(x^2 - 2x + 1) + (y^2 + 4y + 4) = 25$

$x^2 + y^2 - 2x + 4y + 5 = 25$

$x^2 + y^2 - 2x + 4y + 5 - 25 = 0$

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

Both forms $(x - 1)^2 + (y + 2)^2 = 25$ and $x^2 + y^2 - 2x + 4y - 20 = 0$ are valid equations for the circle.

Given:

The equation of the circle is $x^2 + y^2 = 16$.

The equation of the line is $y = \sqrt{3}x + k$.

The line touches the circle (i.e., is tangent to the circle).

To Find:

The value of $k$.

Solution:

The equation of the circle is $x^2 + y^2 = 16$. This is in the standard form $(x - h)^2 + (y - k)^2 = r^2$, where the center is $(h, k) = (0, 0)$ and the radius $r^2 = 16$, so $r = \sqrt{16} = 4$.

The equation of the line is $y = \sqrt{3}x + k$. We can rewrite this equation in the general form $Ax + By + C = 0$:

$\sqrt{3}x - y + k = 0$

Here, $A = \sqrt{3}$, $B = -1$, and $C = k$.

For a line to touch a circle, the perpendicular distance from the center of the circle to the line must be equal to the radius of the circle.

The distance from a point $(x_0, y_0)$ to a line $Ax + By + C = 0$ is given by the formula:

$d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$

The center of the circle is $(0, 0)$, so $(x_0, y_0) = (0, 0)$. The radius is $r = 4$.

Set the distance $d$ equal to the radius $r$:

$r = \frac{|A(0) + B(0) + C|}{\sqrt{A^2 + B^2}}$

$4 = \frac{|\sqrt{3}(0) + (-1)(0) + k|}{\sqrt{(\sqrt{3})^2 + (-1)^2}}$

$4 = \frac{|0 + 0 + k|}{\sqrt{3 + 1}}$

$4 = \frac{|k|}{\sqrt{4}}$

$4 = \frac{|k|}{2}$

Multiply both sides by 2:

$4 \times 2 = |k|$

$8 = |k|$

This means that $k$ can be either 8 or -8.

Therefore, the value of $k$ is $\pm 8$.

Given:

Equation of the first circle: $x^2 + y^2 – 6x + 12y + 15 = 0$.

A second circle is concentric with the first circle and has double its area.

To Find:

The equation of the second circle.

Solution:

The standard form of a circle's equation is $x^2 + y^2 + 2gx + 2fy + c = 0$. The center of this circle is $(-g, -f)$ and the radius is $r = \sqrt{g^2 + f^2 - c}$.

For the given circle $x^2 + y^2 – 6x + 12y + 15 = 0$, we compare it with the general form:

$2g = -6 \implies g = -3$

$2f = 12 \implies f = 6$

$c = 15$

The center of the first circle is $(-g, -f) = (-(-3), -(6)) = (3, -6)$.

The radius squared of the first circle is $r_1^2 = g^2 + f^2 - c = (-3)^2 + (6)^2 - 15 = 9 + 36 - 15 = 45 - 15 = 30$.

The area of the first circle is $A_1 = \pi r_1^2 = \pi (30) = 30\pi$.

The second circle is concentric with the first circle, which means they have the same center. So, the center of the second circle is also $(3, -6)$.

The area of the second circle is double the area of the first circle.

Area of second circle $A_2 = 2 \times A_1 = 2 \times 30\pi = 60\pi$.

Let $r_2$ be the radius of the second circle. The area of the second circle is $A_2 = \pi r_2^2$.

$\pi r_2^2 = 60\pi$

Divide both sides by $\pi$:

$r_2^2 = 60$

The equation of the second circle with center $(h, k) = (3, -6)$ and radius squared $r_2^2 = 60$ is given by the standard form $(x - h)^2 + (y - k)^2 = r^2$:

$(x - 3)^2 + (y - (-6))^2 = 60$

$(x - 3)^2 + (y + 6)^2 = 60$

This is the equation of the second circle.

Expanding the equation, we get:

$(x^2 - 6x + 9) + (y^2 + 12y + 36) = 60$

$x^2 + y^2 - 6x + 12y + 45 = 60$

Subtract 60 from both sides:

$x^2 + y^2 - 6x + 12y + 45 - 60 = 0$

$x^2 + y^2 - 6x + 12y - 15 = 0$

Both forms $(x - 3)^2 + (y + 6)^2 = 60$ and $x^2 + y^2 - 6x + 12y - 15 = 0$ are valid equations for the circle.

Given:

The length of the latus rectum of an ellipse is equal to half of the length of its minor axis.

To Find:

The eccentricity of the ellipse.

Solution:

Let the standard equation of the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.

Let $a$ be the length of the semi-major axis and $b$ be the length of the semi-minor axis.

We assume that the major axis is along the x-axis, so $a > b$. If $b > a$, the derivation for eccentricity will yield the same result.

The length of the minor axis is $2b$.

The length of the latus rectum of the ellipse (with major axis along x-axis) is given by the formula $\frac{2b^2}{a}$.

According to the given condition:

Length of Latus Rectum $= \frac{1}{2} \times$ Length of Minor Axis

Substitute the formulas:

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

Simplify the right side of the equation:

$\frac{2b^2}{a} = b$

Since $b$ represents the length of the semi-minor axis, $b \neq 0$. We can divide both sides of the equation by $b$:

$\frac{2b}{a} = 1$

Now, multiply both sides by $a$:

$2b = a$

Square both sides of this equation:

$(2b)^2 = a^2$

$4b^2 = a^2$

The eccentricity $e$ of an ellipse is related to the semi-major axis $a$ and semi-minor axis $b$ by the equation $b^2 = a^2(1 - e^2)$ when $a > b$.

We need to find the value of $e$. Let's substitute the expression for $b^2$ from the eccentricity relation into the equation $4b^2 = a^2$:

$4 (a^2(1 - e^2)) = a^2$

$4a^2(1 - e^2) = a^2$

Since $a$ is the semi-major axis length, $a \neq 0$. We can divide both sides of the equation by $a^2$:

$4(1 - e^2) = 1$

Divide both sides by 4:

$1 - e^2 = \frac{1}{4}$

Now, rearrange the equation to solve for $e^2$:

$e^2 = 1 - \frac{1}{4}$

$e^2 = \frac{4}{4} - \frac{1}{4}$

$e^2 = \frac{3}{4}$

Take the square root of both sides. Eccentricity $e$ for an ellipse is always a positive value:

$e = \sqrt{\frac{3}{4}}$

$e = \frac{\sqrt{3}}{\sqrt{4}}$

$e = \frac{\sqrt{3}}{2}$

The eccentricity of the ellipse is $\frac{\sqrt{3}}{2}$. This value is between 0 and 1, which is consistent for an ellipse.

Given:

The equation of the ellipse is $9x^2 + 25y^2 = 225$.

To Find:

The eccentricity and the coordinates of the foci of the ellipse.

Solution:

The given equation of the ellipse is $9x^2 + 25y^2 = 225$.

To find the eccentricity and foci, we first need to convert the equation into the standard form of an ellipse centered at the origin, which is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ or $\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$.

Divide the given equation by 225:

$\frac{9x^2}{225} + \frac{25y^2}{225} = \frac{225}{225}$

Simplify the fractions:

$\frac{x^2}{\frac{225}{9}} + \frac{y^2}{\frac{225}{25}} = 1$

$\frac{x^2}{25} + \frac{y^2}{9} = 1$

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

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

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

Since $a^2 > b^2$ ($25 > 9$), the major axis of the ellipse is along the x-axis, and the length of the semi-major axis is $a = 5$, and the length of the semi-minor axis is $b = 3$. The center of the ellipse is at the origin $(0, 0)$.

The eccentricity $e$ of an ellipse (with major axis along the x-axis) is given by the formula:

$e = \sqrt{1 - \frac{b^2}{a^2}}$

Substitute the values of $a^2$ and $b^2$:

$e = \sqrt{1 - \frac{9}{25}}$

$e = \sqrt{\frac{25 - 9}{25}}$

$e = \sqrt{\frac{16}{25}}$

$e = \frac{\sqrt{16}}{\sqrt{25}}$

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

The eccentricity of the ellipse is $\frac{4}{5}$.

The coordinates of the foci are $(\pm c, 0)$, where $c$ is the distance from the center to the foci. For an ellipse with the major axis along the x-axis, $c$ is related to $a$ and $e$ by $c = ae$.

$c = a \times e$

$c = 5 \times \frac{4}{5}$

$c = \cancel{5} \times \frac{4}{\cancel{5}}$

$c = 4$

The distance from the center to the foci is 4.

Since the center is $(0, 0)$ and the major axis is along the x-axis, the coordinates of the foci are $(\pm 4, 0)$.

Thus, the foci are at $(4, 0)$ and $(-4, 0)$.

Given:

The eccentricity of the ellipse is $e = \frac{5}{8}$.

The distance between its foci is 10.

To Find:

The length of the latus rectum of the ellipse.

Solution:

Let the distance from the center to each focus be $c$. The distance between the two foci is $2c$.

We are given that the distance between the foci is 10.

$2c = 10$

Divide both sides by 2:

$c = 5$

The eccentricity $e$ of an ellipse is related to the semi-major axis $a$ and the distance from the center to the focus $c$ by the formula $e = \frac{c}{a}$.

We are given $e = \frac{5}{8}$ and we found $c = 5$. Substitute these values into the formula:

$\frac{5}{8} = \frac{5}{a}$

To solve for $a$, we can cross-multiply or notice that the numerators are equal, so the denominators must be equal:

$5a = 5 \times 8$

$5a = 40$

Divide both sides by 5:

$a = \frac{40}{5}$

$a = 8$

The length of the semi-major axis is $a = 8$.

Now, we need to find the length of the semi-minor axis $b$. The relationship between $a$, $b$, and $e$ for an ellipse is $b^2 = a^2(1 - e^2)$.

Substitute the values of $a = 8$ and $e = \frac{5}{8}$:

$b^2 = 8^2 \left(1 - \left(\frac{5}{8}\right)^2\right)$

$b^2 = 64 \left(1 - \frac{25}{64}\right)$

$b^2 = 64 \left(\frac{64 - 25}{64}\right)$

$b^2 = 64 \left(\frac{39}{64}\right)$

$b^2 = 39$

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

Substitute the values of $b^2 = 39$ and $a = 8$ into the formula:

Latus Rectum $= \frac{2 \times 39}{8}$

Latus Rectum $= \frac{78}{8}$

Simplify the fraction by dividing the numerator and denominator by their greatest common divisor, which is 2:

Latus Rectum $= \frac{\cancel{78}^{39}}{\cancel{8}_{4}}$

Latus Rectum $= \frac{39}{4}$

The length of the latus rectum of the ellipse is $\frac{39}{4}$.

Given:

Eccentricity of the ellipse, $e = \frac{2}{3}$.

Length of the latus rectum $= 5$.

Center of the ellipse is $(0, 0)$.

To Find:

The equation of the ellipse.

Solution:

Let the equation of the ellipse centered at the origin be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. There are two possible cases depending on whether the major axis is along the x-axis or the y-axis.

Let $A$ be the length of the semi-major axis and $B$ be the length of the semi-minor axis. The relationship between the semi-major axis $A$, semi-minor axis $B$, and eccentricity $e$ is given by $B^2 = A^2(1 - e^2)$.

The length of the latus rectum is given by $\frac{2B^2}{A}$.

We are given $e = \frac{2}{3}$ and Latus Rectum $= 5$.

First, calculate $e^2$:

$e^2 = \left(\frac{2}{3}\right)^2 = \frac{4}{9}$

Now, use the relation $B^2 = A^2(1 - e^2)$:

$B^2 = A^2\left(1 - \frac{4}{9}\right)$

$B^2 = A^2\left(\frac{9 - 4}{9}\right)$

$B^2 = \frac{5}{9}A^2$

Next, use the formula for the latus rectum:

$\frac{2B^2}{A} = 5$

Substitute the expression for $B^2$ from the eccentricity relation into the latus rectum equation:

$\frac{2\left(\frac{5}{9}A^2\right)}{A} = 5$

$\frac{\frac{10A^2}{9}}{A} = 5$

$\frac{10A^2}{9A} = 5$

Simplify by canceling $A$ (since $A \neq 0$):

$\frac{10A}{9} = 5$

Solve for $A$:

$10A = 5 \times 9$

$10A = 45$

$A = \frac{45}{10} = \frac{9}{2}$

So, the length of the semi-major axis is $A = \frac{9}{2}$. The square of the semi-major axis is $A^2 = \left(\frac{9}{2}\right)^2 = \frac{81}{4}$.

Now, find the value of $B^2$ using $B^2 = \frac{5}{9}A^2$:

$B^2 = \frac{5}{9} \times \frac{81}{4}$

$B^2 = \frac{5 \times \cancel{81}^{9}}{\cancel{9}_{1} \times 4}$

$B^2 = \frac{5 \times 9}{4} = \frac{45}{4}$

So, the square of the semi-minor axis is $B^2 = \frac{45}{4}$.

We have $A^2 = \frac{81}{4}$ and $B^2 = \frac{45}{4}$. Since $A^2 > B^2$, the semi-major axis squared is $A^2 = \frac{81}{4}$ and the semi-minor axis squared is $B^2 = \frac{45}{4}$.

Case 1: Major axis is along the x-axis.

In this case, the equation of the ellipse centered at the origin is $\frac{x^2}{A^2} + \frac{y^2}{B^2} = 1$, where $A^2 = \frac{81}{4}$ is under the $x^2$ term and $B^2 = \frac{45}{4}$ is under the $y^2$ term.

Substituting the values of $A^2$ and $B^2$:

$\frac{x^2}{81/4} + \frac{y^2}{45/4} = 1$

Inverting the denominators, we get:

$\frac{4x^2}{81} + \frac{4y^2}{45} = 1$

Case 2: Major axis is along the y-axis.

In this case, the equation of the ellipse centered at the origin is $\frac{x^2}{B^2} + \frac{y^2}{A^2} = 1$, where $B^2 = \frac{45}{4}$ is under the $x^2$ term and $A^2 = \frac{81}{4}$ is under the $y^2$ term.

Substituting the values of $B^2$ and $A^2$:

$\frac{x^2}{45/4} + \frac{y^2}{81/4} = 1$

Inverting the denominators, we get:

$\frac{4x^2}{45} + \frac{4y^2}{81} = 1$

Since the problem does not specify the orientation of the major axis, there are two possible equations for the ellipse that satisfy the given conditions:

Equation 1: $\frac{4x^2}{81} + \frac{4y^2}{45} = 1$ (Major axis along x-axis)

Equation 2: $\frac{4x^2}{45} + \frac{4y^2}{81} = 1$ (Major axis along y-axis)

Given:

The equation of the ellipse is $\frac{x^2}{36} + \frac{y^2}{20} = 1$.

To Find:

The distance between the directrices of the ellipse.

Solution:

The given equation of the ellipse is $\frac{x^2}{36} + \frac{y^2}{20} = 1$.

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

Comparing the given equation with the standard form, we have:

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

$b^2 = 20 \implies b = \sqrt{20} = \sqrt{4 \times 5} = 2\sqrt{5}$ (assuming $b > 0$)

Since $a^2 = 36$ is greater than $b^2 = 20$, the major axis of the ellipse lies along the x-axis. The semi-major axis length is $a = 6$ and the semi-minor axis length is $b = \sqrt{20}$.

The eccentricity $e$ of an ellipse with the major axis along the x-axis is given by the formula:

$e = \sqrt{1 - \frac{b^2}{a^2}}$

Substitute the values of $a^2$ and $b^2$:

$e = \sqrt{1 - \frac{20}{36}}$

Simplify the fraction $\frac{20}{36}$ by dividing the numerator and denominator by 4:

$\frac{20}{36} = \frac{\cancel{20}^{5}}{\cancel{36}_{9}} = \frac{5}{9}$

So, the eccentricity is:

$e = \sqrt{1 - \frac{5}{9}}$

$e = \sqrt{\frac{9}{9} - \frac{5}{9}}$

$e = \sqrt{\frac{9 - 5}{9}}$

$e = \sqrt{\frac{4}{9}}$

$e = \frac{\sqrt{4}}{\sqrt{9}} = \frac{2}{3}$

The eccentricity of the ellipse is $\frac{2}{3}$.

For an ellipse with the major axis along the x-axis and centered at the origin, the equations of the directrices are $x = \pm \frac{a}{e}$.

Substitute the values of $a = 6$ and $e = \frac{2}{3}$:

$x = \pm \frac{6}{2/3}$

$x = \pm 6 \times \frac{3}{2}$

$x = \pm \frac{\cancel{6}^{3} \times 3}{\cancel{2}_{1}}$

$x = \pm 3 \times 3$

$x = \pm 9$

The equations of the directrices are $x = 9$ and $x = -9$.

The distance between the two vertical lines $x = x_1$ and $x = x_2$ is $|x_2 - x_1|$.

Distance between directrices $= |9 - (-9)| = |9 + 9| = |18| = 18$.

Alternatively, the distance between the directrices $x = \pm \frac{a}{e}$ is $2 \times \frac{a}{e}$.

Distance $= 2 \times \frac{6}{2/3} = 2 \times 6 \times \frac{3}{2} = \cancel{2} \times 6 \times \frac{3}{\cancel{2}} = 6 \times 3 = 18$.

The distance between the directrices is 18.

Given:

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

The focal distance of a point on the parabola is 4.

To Find:

The coordinates of the point on the parabola.

Solution:

The given equation of the parabola is $y^2 = 8x$. This is in the standard form $y^2 = 4ax$.

Comparing $y^2 = 8x$ with $y^2 = 4ax$, we find $4a = 8$, which gives $a = 2$.

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

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

Let $P(x, y)$ be a point on the parabola whose focal distance is 4.

The focal distance of a point on the parabola is the distance from the point to the focus.

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

We are given that the focal distance is 4.

$\sqrt{(x - 2)^2 + y^2} = 4$

Squaring both sides, we get:

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

Since the point $P(x, y)$ lies on the parabola $y^2 = 8x$, we can substitute $y^2 = 8x$ into the equation:

$(x - 2)^2 + 8x = 16$

Expand $(x - 2)^2$:

$x^2 - 4x + 4 + 8x = 16$

Combine like terms:

$x^2 + 4x + 4 = 16$

Subtract 16 from both sides to form a quadratic equation:

$x^2 + 4x + 4 - 16 = 0$

$x^2 + 4x - 12 = 0$

We can solve this quadratic equation by factoring. We look for two numbers that multiply to -12 and add up to 4. These numbers are 6 and -2.

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

This gives two possible values for $x$: $x = -6$ or $x = 2$.

However, for the parabola $y^2 = 8x$, $y^2$ must be non-negative, which means $8x \ge 0$, so $x \ge 0$. Therefore, $x = -6$ is not a valid x-coordinate for a point on this parabola.

We take the valid value $x = 2$.

Now, substitute $x = 2$ into the equation of the parabola $y^2 = 8x$ to find the corresponding y-values:

$y^2 = 8(2)$

$y^2 = 16$

Taking the square root of both sides:

$y = \pm \sqrt{16}$

$y = \pm 4$

So, the possible y-coordinates are 4 and -4 when $x = 2$.

The coordinates of the points on the parabola whose focal distance is 4 are $(2, 4)$ and $(2, -4)$.

Alternate Solution (using Directrix):

The definition of a parabola is the locus of a point that moves such that its distance from a fixed point (the focus) is equal to its distance from a fixed line (the directrix).

From the equation $y^2 = 8x$, we have $4a = 8$, so $a = 2$.

The focus is $F(2, 0)$.

The equation of the directrix is $x = -a$, which is $x = -2$. We can write the directrix as $x + 2 = 0$.

Let $P(x, y)$ be a point on the parabola. The focal distance of $P$ is given as 4.

According to the definition of the parabola, the focal distance is equal to the perpendicular distance from $P(x, y)$ to the directrix $x + 2 = 0$.

The perpendicular distance from $(x, y)$ to $Ax + By + C = 0$ is $\frac{|Ax + By + C|}{\sqrt{A^2 + B^2}}$.

Distance from $P(x, y)$ to $x + 2 = 0$ is $\frac{|1 \cdot x + 0 \cdot y + 2|}{\sqrt{1^2 + 0^2}} = \frac{|x + 2|}{\sqrt{1}} = |x + 2|$.

We are given that the focal distance is 4, so:

$|x + 2| = 4$

This implies two possibilities:

$x + 2 = 4 \quad$ or $\quad x + 2 = -4$

$x = 4 - 2 \quad$ or $\quad x = -4 - 2$

$x = 2 \quad$ or $\quad x = -6$

As discussed earlier, for the parabola $y^2 = 8x$, the x-coordinate must be non-negative ($x \ge 0$). So, we discard $x = -6$ and take $x = 2$.

Substitute $x = 2$ into the equation of the parabola $y^2 = 8x$:

$y^2 = 8(2)$

$y^2 = 16$

$y = \pm \sqrt{16}$

$y = \pm 4$

The coordinates of the points are $(2, 4)$ and $(2, -4)$.

Given:

The equation of the parabola is $y^2 = 4ax$.

A line segment joins the vertex of the parabola and a point on the parabola.

The line segment makes an angle $\theta$ with the positive x-axis.

To Find:

The length of this line segment.

Solution:

The vertex of the parabola $y^2 = 4ax$ is at the origin $(0, 0)$.

Let $P(x, y)$ be a point on the parabola such that the line segment joining the vertex $V(0, 0)$ and $P(x, y)$ makes an angle $\theta$ with the positive x-axis.

Let the length of the line segment VP be $r$.

Using polar coordinates with the origin as the pole and the positive x-axis as the polar axis, the coordinates of the point $P(x, y)$ can be expressed as:

$x = r \cos \theta$

$y = r \sin \theta$

Since the point $P(x, y)$ lies on the parabola $y^2 = 4ax$, its coordinates must satisfy the equation of the parabola.

Substitute $x = r \cos \theta$ and $y = r \sin \theta$ into the equation $y^2 = 4ax$:

$(r \sin \theta)^2 = 4a (r \cos \theta)$

$r^2 \sin^2 \theta = 4ar \cos \theta$

Rearrange the equation:

$r^2 \sin^2 \theta - 4ar \cos \theta = 0$

Factor out $r$ from the equation:

$r (r \sin^2 \theta - 4a \cos \theta) = 0$

This equation gives two possible values for $r$:

1. $r = 0$: This corresponds to the point $P$ being the vertex $(0, 0)$ itself, where the line segment has zero length. This is one endpoint of the segment.

2. $r \sin^2 \theta - 4a \cos \theta = 0$: This gives the length of the segment to a point $P$ other than the vertex.

Assuming we are looking for the length of the segment to a point distinct from the vertex (i.e., $r \neq 0$), we solve the second part of the equation for $r$:

$r \sin^2 \theta = 4a \cos \theta$

If $\sin^2 \theta \neq 0$, we can divide by $\sin^2 \theta$:

$r = \frac{4a \cos \theta}{\sin^2 \theta}$

Note that $\sin^2 \theta = 0$ when $\theta$ is a multiple of $\pi$ ($0, \pm \pi, \pm 2\pi, ...$). If $\theta = 0$ or $\theta = \pi$, the point lies on the x-axis ($y=0$). Substituting $y=0$ into $y^2 = 4ax$ gives $0 = 4ax$, which implies $x=0$ (since $a \neq 0$). Thus, the only point on the x-axis on the parabola is the vertex $(0,0)$. In this case, the segment length is 0.

If $\cos \theta = 0$ (i.e., $\theta = \pm \pi/2, \pm 3\pi/2, ...$), then $r = \frac{4a \times 0}{\sin^2 \theta} = 0$ (provided $\sin \theta \neq 0$), which again corresponds to the vertex $(0, 0)$. The vertical line $x=0$ intersects $y^2=4ax$ only at $(0,0)$.

Assuming $\theta$ is such that the line segment connects to a point other than the vertex (i.e., $\theta$ is not a multiple of $\pi/2$), the length of the line segment is given by:

$r = \frac{4a \cos \theta}{\sin^2 \theta}$

This can also be written as:

$r = 4a \frac{\cos \theta}{\sin \theta} \cdot \frac{1}{\sin \theta} = 4a \cot \theta \cdot \text{cosec } \theta$

The length of the line segment joining the vertex and the point on the parabola is $\frac{4a \cos \theta}{\sin^2 \theta}$, for angles $\theta$ where this expression is defined and non-zero.

Given:

The vertex of the parabola is $V(0, 4)$.

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

To Find:

The equation of the parabola.

Solution:

The vertex is $V(0, 4)$ and the focus is $F(0, 2)$.

Observe the coordinates of the vertex and the focus. The x-coordinates are the same (both 0), and the y-coordinates are different. This indicates that the axis of symmetry of the parabola is a vertical line ($x=0$, which is the y-axis).

Since the axis of symmetry is vertical, the standard equation of the parabola is of the form $(x - h)^2 = 4a(y - k)$, where $(h, k)$ is the vertex and $a$ is the distance from the vertex to the focus (and from the vertex to the directrix).

The vertex is $(h, k) = (0, 4)$.

The focus is $(h, k + a)$ for a parabola opening upwards ($a > 0$) or $(h, k - a)$ for a parabola opening downwards ($a > 0$). The distance between the vertex $(0, 4)$ and the focus $(0, 2)$ is $|4 - 2| = |2| = 2$. So, the absolute value of $a$ is 2, i.e., $|a|=2$. Thus, $a = \pm 2$.

Since the vertex is at $(0, 4)$ and the focus is at $(0, 2)$, the focus is below the vertex. This means the parabola opens downwards. For a downward opening parabola of the form $(x - h)^2 = 4a(y - k)$, the value of $a$ in the equation is negative. The distance from the vertex to the focus is $|a|$. So, $|a| = 2$. Since it opens downwards, $4a$ must be negative, so $a$ must be negative. Thus, $a = -2$. (Note: in the definition of the form $(x-h)^2 = 4a(y-k)$, $a$ is often taken as the directed distance from the vertex to the focus along the axis, so $a = (y_{focus} - y_{vertex})/1$ in this case, which is $(2-4)/1 = -2$. Alternatively, if we define $a$ as the absolute distance, we use $(x-h)^2 = -4a(y-k)$ with $a=2$. Both approaches lead to the same equation).

Using the standard form $(x - h)^2 = 4a(y - k)$ with $(h, k) = (0, 4)$ and $a = -2$:

$(x - 0)^2 = 4(-2)(y - 4)$

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

$x^2 = -8y + 32$

The equation of the parabola is $x^2 = -8y + 32$.

We can also write it as $x^2 + 8y - 32 = 0$ or $8y = 32 - x^2$ or $y = 4 - \frac{1}{8}x^2$.

Verification using the definition of a parabola:

A parabola is the locus of points equidistant from the focus and the directrix.

The vertex is the midpoint of the segment connecting the focus and the point on the directrix that lies on the axis of symmetry.

The axis of symmetry is the line passing through the vertex and the focus, which is $x = 0$ (the y-axis).

The focus is $F(0, 2)$. The vertex is $V(0, 4)$.

Let the directrix be a horizontal line $y = d$. The axis of symmetry is the line $x=0$. The point where the axis of symmetry intersects the directrix is $(0, d)$.

The vertex $(0, 4)$ is the midpoint of the segment between the focus $(0, 2)$ and the point on the directrix $(0, d)$.

Using the midpoint formula for the y-coordinate:

$4 = \frac{2 + d}{2}$

Multiply by 2:

$8 = 2 + d$

$d = 8 - 2 = 6$

So, the equation of the directrix is $y = 6$, or $y - 6 = 0$.

Let $P(x, y)$ be a point on the parabola. The distance from $P$ to the focus $F(0, 2)$ is $PF = \sqrt{(x - 0)^2 + (y - 2)^2} = \sqrt{x^2 + (y - 2)^2}$.

The distance from $P$ to the directrix $y - 6 = 0$ is the perpendicular distance, which is $\frac{|1 \cdot y - 6|}{\sqrt{0^2 + 1^2}} = |y - 6|$.

According to the definition of the parabola, $PF = |y - 6|$.

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

Squaring both sides:

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

Expand both sides:

$x^2 + (y^2 - 4y + 4) = (y^2 - 12y + 36)$

$x^2 + y^2 - 4y + 4 = y^2 - 12y + 36$

Subtract $y^2$ from both sides:

$x^2 - 4y + 4 = -12y + 36$

Move the terms involving $y$ and constants to the right side:

$x^2 = -12y + 36 + 4y - 4$

$x^2 = (-12y + 4y) + (36 - 4)$

$x^2 = -8y + 32$

This matches the equation obtained using the standard form. The equation of the parabola is $x^2 = -8y + 32$.

Given:

The equation of the line is $y = mx + 1$.

The equation of the parabola is $y^2 = 4x$.

The line is tangent to the parabola.

To Find:

The value of $m$.

Solution:

We are given the equation of the parabola $y^2 = 4x$ and the equation of the line $y = mx + 1$.

For the line to be tangent to the parabola, there must be exactly one point of intersection between the line and the parabola. We can find the intersection points by substituting the expression for $y$ from the line equation into the parabola equation.

Substitute $y = mx + 1$ into $y^2 = 4x$:

$(mx + 1)^2 = 4x$

Expand the left side:

$(mx)^2 + 2(mx)(1) + 1^2 = 4x$

$m^2x^2 + 2mx + 1 = 4x$

Rearrange the terms to form a quadratic equation in $x$:

$m^2x^2 + 2mx - 4x + 1 = 0$

$m^2x^2 + (2m - 4)x + 1 = 0$

This is a quadratic equation in the form $Ax^2 + Bx + C = 0$, where $A = m^2$, $B = 2m - 4$, and $C = 1$.

For the line to be tangent to the parabola, the quadratic equation must have exactly one solution for $x$. This happens when the discriminant of the quadratic equation is zero.

The discriminant is given by $\Delta = B^2 - 4AC$.

Set the discriminant equal to zero:

$B^2 - 4AC = 0$

Substitute the values of $A$, $B$, and $C$:

$(2m - 4)^2 - 4(m^2)(1) = 0$

Expand $(2m - 4)^2$:

$(2m)^2 - 2(2m)(4) + 4^2 - 4m^2 = 0$

$4m^2 - 16m + 16 - 4m^2 = 0$

Combine like terms:

$(4m^2 - 4m^2) - 16m + 16 = 0$

$0 - 16m + 16 = 0$

$-16m + 16 = 0$

Add $16m$ to both sides:

$16 = 16m$

Divide both sides by 16:

$m = \frac{16}{16}$

$m = 1$

The value of $m$ is 1.

Alternate Solution (using Tangency Condition):

For the parabola $y^2 = 4ax$, the condition for the line $y = mx + c'$ to be tangent is $c' = \frac{a}{m}$, provided $m \neq 0$.

The given parabola is $y^2 = 4x$. Comparing this with $y^2 = 4ax$, we get $4a = 4$, so $a = 1$.

The given line is $y = mx + 1$. Comparing this with $y = mx + c'$, we get $c' = 1$.

Using the tangency condition $c' = \frac{a}{m}$:

$1 = \frac{1}{m}$

Multiply both sides by $m$ (assuming $m \neq 0$. If $m=0$, the line is $y=1$, which intersects $y^2=4x$ at $1=4x \implies x=1/4$, a single point of intersection, so $m=0$ is a tangent, but it's a horizontal line, tangent at $(1/4, 1)$. Let's check this: if $m=0$, the quadratic equation is $(2(0)-4)x + 1 = 0 \implies -4x+1=0 \implies x=1/4$. This gives a single point. However, the standard tangent condition $c' = a/m$ applies for non-vertical, non-horizontal tangents. Let's trust the discriminant method which is general.).

$m = 1$

The value of $m$ is 1.

Given:

The distance between the foci of a hyperbola is 16.

The eccentricity of the hyperbola is $e = \sqrt{2}$.

To Find:

The equation of the hyperbola (assuming the center is at the origin and the transverse axis is along one of the coordinate axes).

Solution:

Let the distance from the center of the hyperbola to each focus be $c$. The distance between the two foci is $2c$.

We are given that the distance between the foci is 16.

$2c = 16$

Divide both sides by 2:

$c = 8$

The eccentricity $e$ of a hyperbola is related to the semi-transverse axis $a$ and the distance from the center to the focus $c$ by the formula $e = \frac{c}{a}$.

We are given $e = \sqrt{2}$ and we found $c = 8$. Substitute these values into the formula:

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

Solve for $a$:

$a\sqrt{2} = 8$

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

Rationalize the denominator:

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

The length of the semi-transverse axis is $a = 4\sqrt{2}$. The square of the semi-transverse axis is $a^2 = (4\sqrt{2})^2 = 4^2 \times (\sqrt{2})^2 = 16 \times 2 = 32$.

For a hyperbola, the relationship between $a$, $b$ (semi-conjugate axis), and $c$ is $c^2 = a^2 + b^2$.

We have $c = 8$, so $c^2 = 8^2 = 64$. We have $a^2 = 32$. We need to find $b^2$.

$c^2 = a^2 + b^2$

$64 = 32 + b^2$

Subtract 32 from both sides:

$b^2 = 64 - 32$

$b^2 = 32$

The square of the semi-conjugate axis is $b^2 = 32$.

We assume the center of the hyperbola is at the origin $(0, 0)$. There are two standard forms for a hyperbola centered at the origin:

1. Transverse axis along the x-axis: $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$

2. Transverse axis along the y-axis: $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$

Since the problem does not specify the orientation of the transverse axis, both cases are possible.

Case 1: Transverse axis along the x-axis.

The equation is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.

Substitute $a^2 = 32$ and $b^2 = 32$:

$\frac{x^2}{32} - \frac{y^2}{32} = 1$

Multiply both sides by 32:

$x^2 - y^2 = 32$

Case 2: Transverse axis along the y-axis.

The equation is $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$.

Substitute $a^2 = 32$ and $b^2 = 32$:

$\frac{y^2}{32} - \frac{x^2}{32} = 1$

Multiply both sides by 32:

$y^2 - x^2 = 32$

Both $x^2 - y^2 = 32$ and $y^2 - x^2 = 32$ are possible equations for the hyperbola that satisfy the given conditions.

Note that since $a^2 = b^2$, this hyperbola is a rectangular hyperbola. The asymptotes are $y = \pm \frac{b}{a}x = \pm \sqrt{\frac{32}{32}}x = \pm x$. For the first case, the asymptotes are $y = \pm x$, and for the second case, the asymptotes are $y = \pm x$.

Given:

The equation of the hyperbola is $9y^2 – 4x^2 = 36$.

To Find:

The eccentricity of the hyperbola.

Solution:

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

To find the eccentricity, we need to convert the equation into the standard form of a hyperbola. The standard forms centered at the origin are $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ or $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$. The term with the positive coefficient determines the orientation of the transverse axis, and the denominator of the positive term is $a^2$, while the denominator of the negative term is $b^2$.

Divide the given equation by 36 to make the right side equal to 1:

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

Simplify the fractions:

$\frac{y^2}{4} - \frac{x^2}{9} = 1$

Comparing this equation with the standard form $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$ (since the $y^2$ term is positive), we identify $a^2$ and $b^2$:

$a^2 = 4$

$b^2 = 9$

Here, $a^2$ is under the positive $y^2$ term, so the transverse axis is along the y-axis. The length of the semi-transverse axis squared is $a^2 = 4$, and the length of the semi-conjugate axis squared is $b^2 = 9$.

The eccentricity $e$ of a hyperbola is given by the formula:

$e = \sqrt{1 + \frac{b^2}{a^2}}$

Substitute the values of $a^2 = 4$ and $b^2 = 9$ into the formula:

$e = \sqrt{1 + \frac{9}{4}}$

$e = \sqrt{\frac{4}{4} + \frac{9}{4}}$

$e = \sqrt{\frac{4 + 9}{4}}$

$e = \sqrt{\frac{13}{4}}$

$e = \frac{\sqrt{13}}{\sqrt{4}}$

$e = \frac{\sqrt{13}}{2}$

The eccentricity of the hyperbola is $\frac{\sqrt{13}}{2}$.

Given:

The eccentricity of the hyperbola is $e = \frac{3}{2}$.

The coordinates of the foci are $(\pm 2, 0)$.

To Find:

The equation of the hyperbola.

Solution:

The foci of the hyperbola are given as $(\pm 2, 0)$.

The center of the hyperbola is the midpoint of the segment joining the foci. The midpoint of $(2, 0)$ and $(-2, 0)$ is $\left(\frac{2 + (-2)}{2}, \frac{0 + 0}{2}\right) = (0, 0)$.

So, the center of the hyperbola is at the origin $(0, 0)$.

Since the foci are on the x-axis, the transverse axis of the hyperbola is along the x-axis.

The coordinates of the foci of a hyperbola with center at the origin and transverse axis along the x-axis are $(\pm c, 0)$, where $c$ is the distance from the center to the focus.

Comparing $(\pm c, 0)$ with $(\pm 2, 0)$, we find $c = 2$.

The eccentricity $e$ of a hyperbola is defined as the ratio of the distance from the center to the focus ($c$) to the length of the semi-transverse axis ($a$).

$e = \frac{c}{a}$

We are given $e = \frac{3}{2}$ and we found $c = 2$. Substitute these values:

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

To solve for $a$, we can cross-multiply:

$3a = 2 \times 2$

$3a = 4$

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

The length of the semi-transverse axis is $a = \frac{4}{3}$. The square of the semi-transverse axis is $a^2 = \left(\frac{4}{3}\right)^2 = \frac{16}{9}$.

For a hyperbola, the relationship between $a$ (semi-transverse axis), $b$ (semi-conjugate axis), and $c$ (distance from center to focus) is given by $c^2 = a^2 + b^2$.

We have $c = 2$, so $c^2 = 2^2 = 4$. We have $a^2 = \frac{16}{9}$. We need to find $b^2$.

$4 = \frac{16}{9} + b^2$

Subtract $\frac{16}{9}$ from both sides to find $b^2$:

$b^2 = 4 - \frac{16}{9}$

$b^2 = \frac{4 \times 9}{9} - \frac{16}{9}$

$b^2 = \frac{36}{9} - \frac{16}{9}$

$b^2 = \frac{36 - 16}{9}$

$b^2 = \frac{20}{9}$

The square of the semi-conjugate axis is $b^2 = \frac{20}{9}$.

Since the center is at the origin $(0, 0)$ and the transverse axis is along the x-axis, the standard equation of the hyperbola is:

$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$

Substitute the values of $a^2 = \frac{16}{9}$ and $b^2 = \frac{20}{9}$ into the equation:

$\frac{x^2}{16/9} - \frac{y^2}{20/9} = 1$

We can rewrite this by inverting the denominators:

$\frac{9x^2}{16} - \frac{9y^2}{20} = 1$

This is the equation of the hyperbola.

Given:

Equations of two diameters of a circle are $2x - 3y = 5$ and $3x - 4y = 7$.

Area of the circle = 154 square units.

To Find:

The equation of the circle.

Solution:

The center of the circle is the point of intersection of its diameters. We need to solve the system of linear equations:

$2x - 3y = 5$

$3x - 4y = 7$

Multiply the first equation by 3 and the second equation by 2 to make the coefficients of $x$ equal.

Subtract equation (ii) from equation (i):

Substitute the value of $y = -1$ into the equation $2x - 3y = 5$:

So, the coordinates of the center of the circle are $(1, -1)$. Let the center be $(h, k) = (1, -1)$.

Now, we find the radius of the circle from its area. The area of a circle with radius $r$ is given by the formula $\text{Area} = \pi r^2$.

Given Area = 154 square units. Using $\pi = \frac{22}{7}$:

The radius of the circle is $r = 7$ units.

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

Substitute the values of the center $(h, k) = (1, -1)$ and radius $r = 7$:

This is the required equation of the circle.

Final Answer:

The equation of the circle is $(x - 1)^2 + (y + 1)^2 = 49$.

Given:

The circle passes through points A(2, 3) and B(4, 5).

The center of the circle lies on the line $y - 4x + 3 = 0$.

To Find:

The equation of the circle.

Solution:

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

Since the center $(h, k)$ lies on the line $y - 4x + 3 = 0$, it must satisfy the equation of the line.

The distance from the center to any point on the circle is the radius. Therefore, the distance from $(h, k)$ to A(2, 3) must be equal to the distance from $(h, k)$ to B(4, 5).

Let $r$ be the radius of the circle. Using the distance formula, we have:

Equating the two expressions for $r^2$:

Expand the squares:

Simplify by canceling $h^2$ and $k^2$ from both sides:

Rearrange the terms to group $h$ and $k$:

Divide by 4:

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

Substitute the expression for $k$ from equation (i) into equation (ii):

Now substitute the value of $h = 2$ back into equation (ii) to find $k$:

The center of the circle is $(h, k) = (2, 5)$.

Now, calculate the square of the radius, $r^2$, using the distance from the center $(2, 5)$ to one of the given points, say A(2, 3):

The equation of a circle with center $(h, k)$ and radius $r$ is given by $(x - h)^2 + (y - k)^2 = r^2$.

Substitute the values $h = 2$, $k = 5$, and $r^2 = 4$:

Final Answer:

The equation of the circle is $(x - 2)^2 + (y - 5)^2 = 4$.

Given:

Centre of the circle C = (3, –1).

Length of the chord = 6 units.

Equation of the line containing the chord: $2x – 5y + 18 = 0$.

To Find:

The equation of the circle.

Solution:

Let the center of the circle be C = $(h, k) = (3, -1)$.

Let the line be L: $2x - 5y + 18 = 0$.

Let the chord cut off by the circle on the line L be AB. The length of the chord AB is 6 units.

Let M be the midpoint of the chord AB. The line segment CM, which is the perpendicular distance from the center to the chord, bisects the chord.

So, AM = MB = $\frac{\text{Length of chord}}{2} = \frac{6}{2} = 3$ units.

The distance CM is the perpendicular distance from the center C(3, -1) to the line $2x - 5y + 18 = 0$.

The formula for the perpendicular distance from a point $(x_1, y_1)$ to the line $Ax + By + C = 0$ is $d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$.

Here, $(x_1, y_1) = (3, -1)$, $A = 2$, $B = -5$, and $C = 18$.

Let $r$ be the radius of the circle. In the right-angled triangle CMA, AC is the hypotenuse (radius), AM is half the chord length, and CM is the perpendicular distance.

By the Pythagorean theorem in $\triangle$CMA:

The equation of a circle with center $(h, k)$ and radius $r$ is $(x - h)^2 + (y - k)^2 = r^2$.

Substitute the center $(h, k) = (3, -1)$ and $r^2 = 38$:

Final Answer:

The equation of the circle is $(x - 3)^2 + (y + 1)^2 = 38$.

Given:

Equation of the first circle (C1): $x^2 + y^2 – 2x – 4y – 20 = 0$.

Point of tangency P: (5, 5).

Radius of the second circle (C2): $r_2 = 5$ units.

To Find:

The equation of the second circle (C2).

Solution:

The equation of the first circle (C1) is given by $x^2 + y^2 – 2x – 4y – 20 = 0$.

Comparing this with the standard form $x^2 + y^2 + 2gx + 2fy + c = 0$, we have:

The center of the first circle (O1) is $(-g, -f) = (-(-1), -(-2)) = (1, 2)$.

The radius of the first circle ($r_1$) is $\sqrt{g^2 + f^2 - c}$.

So, the first circle (C1) has center O1(1, 2) and radius $r_1 = 5$.

The second circle (C2) has radius $r_2 = 5$, and it touches C1 at the point P(5, 5).

When two circles touch each other, the point of tangency P lies on the straight line joining their centers. Let the center of the second circle (C2) be O2 = $(h, k)$.

The points O1, P, and O2 are collinear.

The distance from O1(1, 2) to P(5, 5) is O1P = $\sqrt{(5-1)^2 + (5-2)^2} = \sqrt{4^2 + 3^2} = \sqrt{16+9} = \sqrt{25} = 5$. This is equal to $r_1$.

The distance from O2(h, k) to P(5, 5) is O2P. This distance must be equal to the radius of C2, $r_2 = 5$. So, O2P = 5.

Since O1, P, and O2 are collinear, and O1P = 5 and O2P = 5, the point of tangency P(5, 5) must be the midpoint of the line segment joining O1(1, 2) and O2(h, k).

Using the midpoint formula, the coordinates of P are $(\frac{1+h}{2}, \frac{2+k}{2})$.

We equate these coordinates to the coordinates of P(5, 5):

Thus, the center of the second circle (C2) is O2(9, 8).

The radius of the second circle is $r_2 = 5$.

The equation of a circle with center $(h, k)$ and radius $r$ is $(x - h)^2 + (y - k)^2 = r^2$.

Substitute the center $(9, 8)$ and radius $r = 5$ into this equation:

This is the equation of the required circle.

Final Answer:

The equation of the circle is $(x - 9)^2 + (y - 8)^2 = 25$.

Given:

The circle passes through the point P(7, 3).

Radius of the circle, $r = 3$ units.

The centre of the circle lies on the line $y = x - 1$.

To Find:

The equation of the circle.

Solution:

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

Since the centre $(h, k)$ lies on the line $y = x - 1$, we have:

The distance from the centre C$(h, k)$ to the point P(7, 3) on the circle is equal to the radius, which is $r = 3$.

Using the distance formula, the square of the distance CP is equal to $r^2$:

Substitute the expression for $k$ from equation (i) into equation (ii):

Expand the squares:

Combine like terms:

Rearrange the equation into standard quadratic form:

Divide the equation by 2:

Factor the quadratic equation:

This gives two possible values for $h$:

Now, find the corresponding values of $k$ using equation (i), $k = h - 1$.

Case 1: If $h = 4$

The centre is $(4, 3)$.

Case 2: If $h = 7$

The centre is $(7, 6)$.

There are two possible circles that satisfy the given conditions.

The equation of a circle with centre $(h, k)$ and radius $r$ is $(x - h)^2 + (y - k)^2 = r^2$. Here $r^2 = 3^2 = 9$.

For the centre $(4, 3)$:

For the centre $(7, 6)$:

Final Answer:

There are two possible equations for the circle:

$(x - 4)^2 + (y - 3)^2 = 9$

or

$(x - 7)^2 + (y - 6)^2 = 9$

(a) Directrix x = 0, focus at (6, 0)

Given:

Focus F = (6, 0)

Directrix equation: $x = 0$

Solution:

The directrix is the y-axis. The focus is on the x-axis to the right of the origin.

The vertex of the parabola is the midpoint of the perpendicular segment from the focus to the directrix. The point on the directrix closest to the focus is (0, 0).

Vertex V = $\left(\frac{6+0}{2}, \frac{0+0}{2}\right) = (3, 0)$.

The parabola opens towards the focus, so it opens to the right.

The distance from the vertex to the focus is $a = |6 - 3| = 3$.

The standard equation of a parabola opening to the right with vertex $(h, k)$ is $(y - k)^2 = 4a(x - h)$.

Substituting $(h, k) = (3, 0)$ and $a = 3$:

(b) Vertex at (0, 4), focus at (0, 2)

Given:

Vertex V = (0, 4)

Focus F = (0, 2)

Solution:

The vertex and focus have the same x-coordinate (0), so the parabola is vertical.

The focus (0, 2) is below the vertex (0, 4), so the parabola opens downwards.

The distance from the vertex to the focus is $a = |4 - 2| = 2$.

The standard equation of a parabola opening downwards with vertex $(h, k)$ is $(x - h)^2 = -4a(y - k)$.

Substituting $(h, k) = (0, 4)$ and $a = 2$:

(c) Focus at (–1, –2), directrix x – 2y + 3 = 0

Given:

Focus F = (–1, –2)

Directrix equation: $x – 2y + 3 = 0$

Solution:

By the definition of a parabola, any point P(x, y) on the parabola is equidistant from the focus and the directrix.

Distance from P(x, y) to the focus F(–1, –2):

Distance from P(x, y) to the directrix $x – 2y + 3 = 0$ (using the formula $\frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$):

According to the definition, PF = PD:

Square both sides:

Expand both sides:

Move all terms to the left side:

Final Answers:

(a) The equation of the parabola is $y^2 = 12(x - 3)$.

(b) The equation of the parabola is $x^2 = -8(y - 4)$.

(c) The equation of the parabola is $4x^2 - 4xy + y^2 + 4x + 32y + 16 = 0$.

Given:

Two fixed points (foci), $F_1 = (3, 0)$ and $F_2 = (9, 0)$.

The sum of the distances from any point on the set to these two points is 12.

To Find:

The equation of the set of all such points.

Solution:

Let P(x, y) be any point in the set. According to the problem statement, the sum of the distances from P to the points (3, 0) and (9, 0) is 12.

This is the definition of an ellipse, where the two fixed points are the foci and the constant sum of distances is equal to the length of the major axis, $2a$.

Given sum of distances = 12.

The foci are $F_1(3, 0)$ and $F_2(9, 0)$. The distance between the foci is $2c$.

For an ellipse, the relationship between $a$, $b$, and $c$ is $a^2 = b^2 + c^2$, where $b$ is the semi-minor axis.

The center of the ellipse is the midpoint of the segment joining the foci.

Center $(h, k) = \left(\frac{3 + 9}{2}, \frac{0 + 0}{2}\right) = \left(\frac{12}{2}, 0\right) = (6, 0)$.

Since the foci lie on the x-axis and the center is $(6, 0)$, the major axis is horizontal.

The standard equation of a horizontal ellipse with center $(h, k)$ is $\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1$.

Substitute the values $h = 6$, $k = 0$, $a^2 = 36$, and $b^2 = 27$:

This is the equation of the set of all points satisfying the given condition.

Final Answer:

The equation of the set of all points is $\frac{(x - 6)^2}{36} + \frac{y^2}{27} = 1$.

Given:

Fixed point (Focus) F = $(0, 4)$.

Fixed line (Directrix) equation: $y = 9$.

Ratio of distances (Eccentricity) $e = \frac{2}{3}$.

To Find:

The equation of the set of all such points.

Solution:

Let P(x, y) be any point on the set. According to the definition of a conic section, the distance from P to the focus is $e$ times the distance from P to the directrix.

Distance from P(x, y) to the focus F(0, 4) is PF:

$PF = \sqrt{(x - 0)^2 + (y - 4)^2} = \sqrt{x^2 + (y - 4)^2}$

The directrix is the line $y = 9$, which can be written as $0x + 1y - 9 = 0$. The perpendicular distance from P(x, y) to this line is PD:

$PD = \frac{|0(x) + 1(y) - 9|}{\sqrt{0^2 + 1^2}} = \frac{|y - 9|}{\sqrt{1}} = |y - 9|$

According to the given condition, $PF = e \times PD$:

$\sqrt{x^2 + (y - 4)^2} = \frac{2}{3} |y - 9|$

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

$(\sqrt{x^2 + (y - 4)^2})^2 = \left(\frac{2}{3} |y - 9|\right)^2$

$x^2 + (y - 4)^2 = \frac{4}{9} (y - 9)^2$

Multiply both sides by 9:

$9[x^2 + (y - 4)^2] = 4(y - 9)^2$

Expand the squared terms:

$9[x^2 + (y^2 - 8y + 16)] = 4[y^2 - 18y + 81]$

$9x^2 + 9y^2 - 72y + 144 = 4y^2 - 72y + 324$

Move all terms to the left side of the equation:

$9x^2 + 9y^2 - 4y^2 - 72y + 72y + 144 - 324 = 0$

Combine the like terms:

$9x^2 + 5y^2 - 180 = 0$

This is the equation of the set of points. It can be written in the standard form of an ellipse by rearranging and dividing by 180:

$9x^2 + 5y^2 = 180$

$\frac{9x^2}{180} + \frac{5y^2}{180} = \frac{180}{180}$

$\frac{x^2}{20} + \frac{y^2}{36} = 1$

Since the eccentricity $e = \frac{2}{3} < 1$, the set of points forms an ellipse.

Final Answer:

The equation of the set of all points is $9x^2 + 5y^2 - 180 = 0$ or $\frac{x^2}{20} + \frac{y^2}{36} = 1$.

Given:

The set of all points P(x, y) such that the absolute difference of their distances from two fixed points F$_1$(4, 0) and F$_2$(–4, 0) is always equal to 2.

This condition is expressed as $|PF_1 - PF_2| = 2$.

To Show:

The set of these points represents a hyperbola.

Solution:

Let the fixed points be the foci F$_1$(4, 0) and F$_2$(–4, 0). Let P(x, y) be any point in the set.

The distance between P(x, y) and F$_1$(4, 0) is given by the distance formula:

$PF_1 = \sqrt{(x - 4)^2 + (y - 0)^2} = \sqrt{(x - 4)^2 + y^2}$

The distance between P(x, y) and F$_2$(–4, 0) is given by the distance formula:

$PF_2 = \sqrt{(x - (-4))^2 + (y - 0)^2} = \sqrt{(x + 4)^2 + y^2}$

According to the given condition, the absolute difference of these distances is 2:

$|\sqrt{(x - 4)^2 + y^2} - \sqrt{(x + 4)^2 + y^2}| = 2$

This means either $\sqrt{(x - 4)^2 + y^2} - \sqrt{(x + 4)^2 + y^2} = 2$ or $\sqrt{(x - 4)^2 + y^2} - \sqrt{(x + 4)^2 + y^2} = -2$.

We can write this as:

$\sqrt{(x - 4)^2 + y^2} = \pm 2 + \sqrt{(x + 4)^2 + y^2}$

Square both sides of the equation:

$((x - 4)^2 + y^2) = (\pm 2 + \sqrt{(x + 4)^2 + y^2})^2$

Expand both sides:

$(x^2 - 8x + 16 + y^2) = (\pm 2)^2 \pm 2(\pm 2)\sqrt{(x + 4)^2 + y^2} + ((x + 4)^2 + y^2)$

$x^2 - 8x + 16 + y^2 = 4 \pm 4\sqrt{(x + 4)^2 + y^2} + (x^2 + 8x + 16 + y^2)$

Cancel the common terms $x^2$, $y^2$, and 16 from both sides:

$-8x = 4 \pm 4\sqrt{(x + 4)^2 + y^2} + 8x$

Isolate the term with the radical:

$-8x - 8x - 4 = \pm 4\sqrt{(x + 4)^2 + y^2}$

$-16x - 4 = \pm 4\sqrt{(x + 4)^2 + y^2}$

Divide both sides by 4:

$-4x - 1 = \pm \sqrt{(x + 4)^2 + y^2}$

Square both sides again to eliminate the radical:

$(-4x - 1)^2 = (\pm \sqrt{(x + 4)^2 + y^2})^2$

$(4x + 1)^2 = (x + 4)^2 + y^2$

Expand the squares:

$16x^2 + 8x + 1 = x^2 + 8x + 16 + y^2$

Cancel the $8x$ term from both sides:

$16x^2 + 1 = x^2 + 16 + y^2$

Rearrange the terms to group the $x^2$ and $y^2$ terms on one side and constant terms on the other:

$16x^2 - x^2 - y^2 = 16 - 1$

$15x^2 - y^2 = 15$

Divide the entire equation by 15 to get the standard form:

$\frac{15x^2}{15} - \frac{y^2}{15} = \frac{15}{15}$

$\frac{x^2}{1} - \frac{y^2}{15} = 1$

This equation is in the standard form of a hyperbola centered at the origin $\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1$, where $(h, k) = (0, 0)$, $a^2 = 1$, and $b^2 = 15$.

The constant difference in distances, 2, is equal to $2a$, so $a = 1$ and $a^2 = 1$.

The distance between the foci (4, 0) and (-4, 0) is $2c = \sqrt{(4 - (-4))^2 + (0 - 0)^2} = \sqrt{8^2} = 8$, so $c = 4$ and $c^2 = 16$.

For a hyperbola, the relationship between $a$, $b$, and $c$ is $c^2 = a^2 + b^2$.

$16 = 1 + b^2 \implies b^2 = 15$. This matches the value found in the equation.

Since the derived equation $\frac{x^2}{1} - \frac{y^2}{15} = 1$ is the standard form of a hyperbola, the set of all points satisfying the given condition represents a hyperbola.

Conclusion:

The equation representing the set of points is $\frac{x^2}{1} - \frac{y^2}{15} = 1$. This is the standard equation of a hyperbola. Hence, the set of all points such that the difference of their distances from (4, 0) and (– 4, 0) is always equal to 2 represents a hyperbola.

Solution for (a):

Given:

Vertices are $(\pm 5, 0)$.

Foci are $(\pm 7, 0)$.

Solution:

Since the vertices and foci lie on the x-axis, the equation of the hyperbola is of the form:

$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$

The vertices are $(\pm a, 0)$.

Thus, $a^2 = 5^2 = 25$.

The foci are $(\pm c, 0)$.

Thus, $c^2 = 7^2 = 49$.

For a hyperbola, we have the relation $c^2 = a^2 + b^2$.

We can find $b^2$ using this relation:

Substituting the values of $a^2$ and $c^2$ into equation (i):

$b^2 = 49 - 25$

$b^2 = 24$

Substituting the values of $a^2$ and $b^2$ into the standard equation of the hyperbola:

$\frac{x^2}{25} - \frac{y^2}{24} = 1$

This is the required equation of the hyperbola.

Solution for (b):

Given:

Vertices are $(0, \pm 7)$.

Eccentricity $e = \frac{4}{3}$.

Solution:

Since the vertices lie on the y-axis, the equation of the hyperbola is of the form:

$\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$

The vertices are $(0, \pm a)$.

Thus, $a^2 = 7^2 = 49$.

The eccentricity of a hyperbola is given by $e = \frac{c}{a}$.

So, $\frac{c}{a} = \frac{4}{3}$.

Substituting $a = 7$:

$\frac{c}{7} = \frac{4}{3}$

$c = 7 \times \frac{4}{3} = \frac{28}{3}$

Thus, $c^2 = \left(\frac{28}{3}\right)^2 = \frac{784}{9}$.

For a hyperbola, we have the relation $c^2 = a^2 + b^2$.

We can find $b^2$ using this relation:

Substituting the values of $a^2$ and $c^2$ into equation (ii):

$b^2 = \frac{784}{9} - 49$

$b^2 = \frac{784 - 49 \times 9}{9}$

$b^2 = \frac{784 - 441}{9}$

$b^2 = \frac{343}{9}$

Substituting the values of $a^2$ and $b^2$ into the standard equation of the hyperbola:

$\frac{y^2}{49} - \frac{x^2}{343/9} = 1$

This can be written as:

$\frac{y^2}{49} - \frac{9x^2}{343} = 1$

This is the required equation of the hyperbola.

Solution for (c):

Given:

Foci are $(0, \pm \sqrt{10})$.

Hyperbola passes through the point $(2, 3)$.

Solution:

Since the foci lie on the y-axis, the equation of the hyperbola is of the form:

$\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$

The foci are $(0, \pm c)$.

Thus, $c^2 = (\sqrt{10})^2 = 10$.

For a hyperbola, we have the relation $c^2 = a^2 + b^2$.

Substituting $c^2 = 10$:

From equation (iii), we can express $b^2$ in terms of $a^2$:

$b^2 = 10 - a^2$

The hyperbola passes through the point $(2, 3)$. Substituting $x=2$ and $y=3$ into the equation of the hyperbola:

$\frac{9}{a^2} - \frac{4}{b^2} = 1$

Substitute $b^2 = 10 - a^2$ into equation (iv):

$\frac{9}{a^2} - \frac{4}{10 - a^2} = 1$

Multiply both sides by $a^2(10 - a^2)$ to clear the denominators:

$9(10 - a^2) - 4a^2 = a^2(10 - a^2)$

$90 - 9a^2 - 4a^2 = 10a^2 - a^4$

$90 - 13a^2 = 10a^2 - a^4$

Rearrange the terms to form a quadratic equation in $a^2$:

$a^4 - 10a^2 - 13a^2 + 90 = 0$

$a^4 - 23a^2 + 90 = 0$

Let $u = a^2$. The equation becomes a quadratic equation in $u$:

$u^2 - 23u + 90 = 0$

We can solve this quadratic equation by factoring. We need two numbers that multiply to 90 and add up to -23. These numbers are -18 and -5.

$(u - 18)(u - 5) = 0$

This gives two possible values for $u$ (which is $a^2$):

$u = 18$ or $u = 5$

So, $a^2 = 18$ or $a^2 = 5$.

Now we find the corresponding values of $b^2$ using $b^2 = 10 - a^2$ (from equation iii):

Case 1: If $a^2 = 18$

$b^2 = 10 - 18 = -8$

Since $b^2$ must be positive for a hyperbola, this case is not possible.

Case 2: If $a^2 = 5$

$b^2 = 10 - 5 = 5$

This is a valid case as $b^2$ is positive.

So, we have $a^2 = 5$ and $b^2 = 5$.

Substitute these values into the standard equation of the hyperbola $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$:

$\frac{y^2}{5} - \frac{x^2}{5} = 1$

This is the required equation of the hyperbola.

The statement is False.

Justification:

For a line to be a diameter of a circle, it must pass through the center of the circle.

The given equation of the circle is $x^2 + y^2 + 6x + 2y = 0$.

The general equation of a circle is $x^2 + y^2 + 2gx + 2fy + c = 0$, where the center is at $(-g, -f)$.

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

$2g = 6 \implies g = 3$

$2f = 2 \implies f = 1$

$c = 0$

The center of the given circle is $(-g, -f) = (-3, -1)$.

Now, we check if this center point $(-3, -1)$ lies on the given line $x + 3y = 0$.

Substitute $x = -3$ and $y = -1$ into the equation of the line:

Left Hand Side (LHS) $= x + 3y = (-3) + 3(-1) = -3 - 3 = -6$

Right Hand Side (RHS) $= 0$

Since LHS $\neq$ RHS ($-6 \neq 0$), the point $(-3, -1)$ does not lie on the line $x + 3y = 0$.

Therefore, the line $x + 3y = 0$ does not pass through the center of the circle.

Hence, the line is not a diameter of the circle.

The statement is False.

Justification:

The shortest distance from a point to a circle depends on whether the point is inside, outside, or on the circle.

The equation of the given circle is $x^2 + y^2 - 14x - 10y - 151 = 0$.

Comparing this with the general equation of a circle $x^2 + y^2 + 2gx + 2fy + c = 0$, we find the center $(C)$ and the radius $(r)$:

$2g = -14 \implies g = -7$

$2f = -10 \implies f = -5$

$c = -151$

The center of the circle is $C(-g, -f) = (7, 5)$.

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

$r = \sqrt{(-7)^2 + (-5)^2 - (-151)}$

$r = \sqrt{49 + 25 + 151}$

$r = \sqrt{225}$

$r = 15$

Let the given point be $P(2, -7)$.

We need to find the distance between the center $C(7, 5)$ and the point $P(2, -7)$. Let this distance be $d$.

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

$d = \sqrt{(2 - 7)^2 + (-7 - 5)^2}$

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

$d = \sqrt{25 + 144}$

$d = \sqrt{169}$

$d = 13$

Now compare the distance $d$ from the center to the point with the radius $r$ of the circle:

$d = 13$ and $r = 15$.

Since $d < r$ ($13 < 15$), the point $(2, -7)$ lies inside the circle.

When a point is inside the circle, the shortest distance from the point to the circle is the difference between the radius and the distance from the center to the point.

Shortest distance $= r - d$

Shortest distance $= 15 - 13$

Shortest distance $= 2$

The calculated shortest distance is 2, which is not equal to 5 as stated in the question.

Therefore, the statement is false.

The statement is True.

Justification:

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

The center of this circle is at the origin $(0, 0)$, and its radius is $r = a$.

The equation of the given line is $lx + my = 1$, which can be written as $lx + my - 1 = 0$.

For a line to be tangent to a circle, the perpendicular distance from the center of the circle to the line must be equal to the radius of the circle.

The formula for the perpendicular distance from a point $(x_0, y_0)$ to the line $Ax + By + C = 0$ is $\frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$.

Here, the point is the center of the circle $(0, 0)$, and the line is $lx + my - 1 = 0$. So, $x_0 = 0$, $y_0 = 0$, $A = l$, $B = m$, and $C = -1$.

The distance from the center $(0, 0)$ to the line $lx + my - 1 = 0$ is:

$d = \frac{|l(0) + m(0) - 1|}{\sqrt{l^2 + m^2}} = \frac{|-1|}{\sqrt{l^2 + m^2}} = \frac{1}{\sqrt{l^2 + m^2}}$

Since the line is a tangent, this distance must be equal to the radius $a$:

Square both sides of equation (i):

$\left(\frac{1}{\sqrt{l^2 + m^2}}\right)^2 = a^2$

$\frac{1}{l^2 + m^2} = a^2$

Rearrange the equation:

$l^2 + m^2 = \frac{1}{a^2}$

This equation describes the locus of the point $(l, m)$. This is the equation of a circle centered at the origin $(0, 0)$ with radius $\sqrt{\frac{1}{a^2}}$. Since $a$ is a radius, $a > 0$, so $\sqrt{\frac{1}{a^2}} = \frac{1}{a}$.

Thus, the point $(l, m)$ lies on the circle $l^2 + m^2 = \left(\frac{1}{a}\right)^2$.

Therefore, the statement is true.

The statement is False.

Justification:

To determine if a point lies inside, outside, or on a circle, we can substitute the coordinates of the point into the equation of the circle and observe the result.

The equation of the given circle is $x^2 + y^2 - 2x + 6y + 1 = 0$.

Let the given point be $(x_1, y_1) = (1, 2)$.

Substitute $x_1 = 1$ and $y_1 = 2$ into the left side of the circle's equation:

Value $= x_1^2 + y_1^2 - 2x_1 + 6y_1 + 1$

Value $= (1)^2 + (2)^2 - 2(1) + 6(2) + 1$

Value $= 1 + 4 - 2 + 12 + 1$

Value $= 5 - 2 + 12 + 1$

Value $= 3 + 13$

Value $= 16$

For a point $(x_1, y_1)$ and a circle $x^2 + y^2 + 2gx + 2fy + c = 0$:

• If $x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c < 0$, the point is inside the circle.
• If $x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c = 0$, the point is on the circle.
• If $x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c > 0$, the point is outside the circle.

In this case, the value is 16, which is greater than 0 ($16 > 0$).

Therefore, the point $(1, 2)$ lies outside the circle $x^2 + y^2 - 2x + 6y + 1 = 0$.

Hence, the statement is false.

The statement is True.

Justification:

The equation of the given parabola is $y^2 = 4ax$.

The equation of the given line is $lx + my + n = 0$.

We need to find the condition under which the line is tangent to the parabola.

There are a few ways to approach this. One common method is to find the points of intersection and impose the condition that there is exactly one point of intersection.

From the equation of the line, if $m \neq 0$, we can write $my = -lx - n$, so $y = -\frac{l}{m}x - \frac{n}{m}$.

Substitute this value of $y$ into the equation of the parabola:

$\left(-\frac{l}{m}x - \frac{n}{m}\right)^2 = 4ax$

$\left(\frac{lx + n}{m}\right)^2 = 4ax$

$\frac{(lx + n)^2}{m^2} = 4ax$

$(lx + n)^2 = 4ax m^2$

$l^2x^2 + 2lnx + n^2 = 4am^2x$

Rearrange into a quadratic equation in $x$:

$l^2x^2 + (2ln - 4am^2)x + n^2 = 0$

For the line to be tangent to the parabola, this quadratic equation must have exactly one solution for $x$. This occurs when the discriminant ($\Delta$) of the quadratic equation is zero.

The discriminant is given by $\Delta = B^2 - 4AC$, where $A = l^2$, $B = 2ln - 4am^2$, and $C = n^2$.

So, we must have $\Delta = (2ln - 4am^2)^2 - 4(l^2)(n^2) = 0$.

$(2ln)^2 - 2(2ln)(4am^2) + (4am^2)^2 - 4l^2n^2 = 0$

$4l^2n^2 - 16alnm^2 + 16a^2m^4 - 4l^2n^2 = 0$

The $4l^2n^2$ terms cancel out:

$-16alnm^2 + 16a^2m^4 = 0$

Divide by $16a$ (assuming $a \neq 0$, otherwise $y^2=0$, which is the x-axis, and the condition needs re-evaluation):

$-lnm^2 + am^4 = 0$

$am^4 - lnm^2 = 0$

$m^2(am^2 - ln) = 0$

This equation gives two possibilities: $m^2 = 0$ or $am^2 - ln = 0$.

Possibility 1: $m^2 = 0 \implies m = 0$.

If $m = 0$, the equation of the line becomes $lx + n = 0$. If $l \neq 0$, this is a vertical line $x = -n/l$. A vertical line is tangent to the parabola $y^2 = 4ax$ (with $a \neq 0$) only if it is the y-axis, i.e., $x = 0$. This means $-n/l = 0$, so $n=0$. In this case ($m=0, n=0, l \neq 0$), the condition $ln = am^2$ becomes $l(0) = a(0)^2$, which is $0 = 0$. This holds true, confirming that the condition works for the vertical tangent at the vertex (if $a \neq 0$).

Possibility 2: $am^2 - ln = 0 \implies ln = am^2$.

This is the condition for tangency when $m \neq 0$ (and also covers the vertical tangent case as shown above).

Thus, the line $lx + my + n = 0$ is tangent to the parabola $y^2 = 4ax$ if and only if $ln = am^2$ (assuming $a \neq 0$ and $l, m$ are not both zero). If $a=0$, the parabola is $y^2=0$, which is the line $y=0$. The line $lx+my+n=0$ is tangent to $y=0$ if it is the same line (i.e., $l=0, n=0$) or if it intersects at one point and is not parallel to the x-axis. If $l=0, n=0$, the line is $my=0$, which is $y=0$ if $m \neq 0$. The condition $ln=am^2$ becomes $0=0$ which holds. If $l \neq 0$, $y=-n/m - (l/m)x$. For tangency to $y=0$, it must be the same line, so $l=0$ and $n=0$. The condition $ln=am^2$ becomes $0=0$ which holds. So the condition holds even for $a=0$ cases where tangency is meaningfully defined.

The condition for the line $lx + my + n = 0$ to be tangent to $y^2 = 4ax$ is indeed $ln = am^2$.

Therefore, the statement is true.

Alternate Justification using standard tangent form:

The equation of a tangent to the parabola $y^2 = 4ax$ in slope form is $y = m'x + \frac{a}{m'}$ (where $m'$ is the slope of the tangent). This form covers all tangents except the vertical tangent $x=0$.

The given line is $lx + my + n = 0$. If $m \neq 0$, we can write this as $my = -lx - n$, or $y = -\frac{l}{m}x - \frac{n}{m}$.

Comparing this with the tangent form $y = m'x + c'$, we have the slope $m' = -\frac{l}{m}$ and the y-intercept $c' = -\frac{n}{m}$.

For the line to be a tangent, the condition $c' = \frac{a}{m'}$ must hold:

$-\frac{n}{m} = \frac{a}{-\frac{l}{m}}$

$-\frac{n}{m} = a \times \left(-\frac{m}{l}\right)$

$-\frac{n}{m} = -\frac{am}{l}$

Multiply both sides by $-lm$ (assuming $l \neq 0$ and $m \neq 0$):

$(-lm)\left(-\frac{n}{m}\right) = (-lm)\left(-\frac{am}{l}\right)$

$ln = am^2$

This condition $ln = am^2$ is derived assuming $m \neq 0$ and $l \neq 0$. As shown in the first justification, this condition also correctly holds for the vertical tangent $x=0$ (where $m=0, n=0, l \neq 0$, assuming $a \neq 0$).

Thus, the condition $ln = am^2$ is the general condition for the line $lx + my + n = 0$ to be tangent to the parabola $y^2 = 4ax$.

The statement is False.

Justification:

The given equation of the ellipse is $\frac{x^2}{16} + \frac{y^2}{25} = 1$.

The standard equation of an ellipse centered at the origin is $\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$ when the major axis is along the y-axis ($a > b$).

Comparing the given equation with the standard form, we have:

$a^2 = 25 \implies a = \sqrt{25} = 5$ (Since $a$ is a length, $a > 0$)

$b^2 = 16 \implies b = \sqrt{16} = 4$ (Since $b$ is a length, $b > 0$)

Since $a^2 > b^2$, the major axis is along the y-axis. The length of the semi-major axis is $a = 5$.

The foci of an ellipse are points S and S' such that for any point P on the ellipse, the sum of the distances PS + PS' is a constant. This constant sum is equal to the length of the major axis, which is $2a$.

Substitute the value of $a$ into equation (i):

PS + PS' = $2 \times 5$

PS + PS' = $10$

The sum of the distances from any point P on the given ellipse to its foci is 10.

The statement claims that PS + PS' = 8.

Since $10 \neq 8$, the statement is false.

The statement is True.

Justification:

For a line to touch an ellipse at a specific point, two conditions must be met:

1. The point must lie on the ellipse.

2. The equation of the tangent to the ellipse at that point must be the same as the equation of the given line.

The equation of the given ellipse is $\frac{x^2}{9} + \frac{y^2}{4} = 2$.

The given point is $(3, 2)$. Let's check if this point lies on the ellipse by substituting $x=3$ and $y=2$ into the ellipse equation:

Left Hand Side (LHS) $= \frac{(3)^2}{9} + \frac{(2)^2}{4} = \frac{9}{9} + \frac{4}{4} = 1 + 1 = 2$.

Right Hand Side (RHS) $= 2$.

Since LHS = RHS, the point $(3, 2)$ lies on the ellipse.

Now, let's find the equation of the tangent to the ellipse at the point $(3, 2)$.

The equation of the ellipse can be rewritten in the standard form $\frac{X^2}{A^2} + \frac{Y^2}{B^2} = 1$.

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

Divide by 2:

$\frac{x^2}{18} + \frac{y^2}{8} = 1$

Here, $A^2 = 18$ and $B^2 = 8$. The point of tangency is $(x_1, y_1) = (3, 2)$.

The equation of the tangent to the ellipse $\frac{x^2}{A^2} + \frac{y^2}{B^2} = 1$ at the point $(x_1, y_1)$ is given by:

Substitute $x_1 = 3$, $y_1 = 2$, $A^2 = 18$, and $B^2 = 8$ into equation (i):

$\frac{x(3)}{18} + \frac{y(2)}{8} = 1$

$\frac{3x}{18} + \frac{2y}{8} = 1$

Simplify the fractions:

$\frac{x}{6} + \frac{y}{4} = 1$

To clear the denominators, multiply the entire equation by the least common multiple (LCM) of 6 and 4, which is 12:

$12 \left(\frac{x}{6}\right) + 12 \left(\frac{y}{4}\right) = 12(1)$

$2x + 3y = 12$

This is the equation of the tangent to the ellipse at the point $(3, 2)$.

The given line is $2x + 3y = 12$.

Since the equation of the tangent at $(3, 2)$ is the same as the equation of the given line, the line $2x + 3y = 12$ touches the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 2$ at the point $(3, 2)$.

Therefore, the statement is true.

The statement is True.

Justification:

We are given the equations of two lines:

Line 1: $\sqrt{3}x - y - 4\sqrt{3}k = 0$

Line 2: $\sqrt{3}kx + ky − 4\sqrt{3} = 0$

We need to find the locus of the point of intersection $(x, y)$ of these two lines by eliminating the parameter $k$.

Rewrite the equations to isolate terms involving $k$:

From Line 1:

$\sqrt{3}x - y = 4\sqrt{3}k$

Assuming $4\sqrt{3} \neq 0$, we can express $k$ as:

From Line 2:

$k(\sqrt{3}x + y) = 4\sqrt{3}$

Assuming $\sqrt{3}x + y \neq 0$, we can express $k$ as:

Since equations (1) and (2) both represent the same parameter $k$, we can equate the expressions:

$\frac{\sqrt{3}x - y}{4\sqrt{3}} = \frac{4\sqrt{3}}{\sqrt{3}x + y}$

Cross-multiply the terms:

$(\sqrt{3}x - y)(\sqrt{3}x + y) = (4\sqrt{3})(4\sqrt{3})$

Using the difference of squares formula $(a-b)(a+b) = a^2 - b^2$ on the left side and simplifying the right side:

$(\sqrt{3}x)^2 - y^2 = (4\sqrt{3})^2$

$3x^2 - y^2 = 16 \times 3$

$3x^2 - y^2 = 48$

To express this equation in the standard form of a conic section, divide the entire equation by 48:

$\frac{3x^2}{48} - \frac{y^2}{48} = \frac{48}{48}$

Equation (3) is of the form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, which represents a hyperbola with the transverse axis along the x-axis.

Comparing equation (3) with the standard form, we find the values of $a^2$ and $b^2$:

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

$b^2 = 48$

The eccentricity ($e$) of a hyperbola with the transverse axis along the x-axis is calculated using the formula:

$e = \sqrt{1 + \frac{b^2}{a^2}}$

Substitute the values of $a^2$ and $b^2$ into the eccentricity formula:

$e = \sqrt{1 + \frac{48}{16}}$

$e = \sqrt{1 + 3}$

$e = \sqrt{4}$

$e = 2$

The locus of the point of intersection of the given lines for different values of $k$ is a hyperbola with an eccentricity of 2.

This result matches the description provided in the statement.

Therefore, the statement is true.

The equation of the circle having centre at $(3, -4)$ and touching the line $5x + 12y – 12 = 0$ is $\mathbf{(x - 3)^2 + (y + 4)^2 = \frac{2025}{169}}$ .

Solution:

The center of the circle is given as $(h, k) = (3, -4)$.

The line $5x + 12y – 12 = 0$ is tangent to the circle.

The radius ($r$) of the circle is the perpendicular distance from the center $(3, -4)$ to the tangent line $5x + 12y – 12 = 0$.

The formula for the perpendicular distance from a point $(x_0, y_0)$ to a line $Ax + By + C = 0$ is $\frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$.

Here, $(x_0, y_0) = (3, -4)$, $A = 5$, $B = 12$, and $C = -12$.

The radius $r$ is:

$r = \frac{|5(3) + 12(-4) - 12|}{\sqrt{5^2 + 12^2}}$

$r = \frac{|15 - 48 - 12|}{\sqrt{25 + 144}}$

$r = \frac{|-45|}{\sqrt{169}}$

$r = \frac{45}{13}$

The square of the radius is $r^2 = \left(\frac{45}{13}\right)^2 = \frac{45^2}{13^2} = \frac{2025}{169}$.

The equation of a circle with center $(h, k)$ and radius $r$ is $(x - h)^2 + (y - k)^2 = r^2$.

Substitute the center $(3, -4)$ and $r^2 = \frac{2025}{169}$ into the equation:

$(x - 3)^2 + (y - (-4))^2 = \frac{2025}{169}$

$(x - 3)^2 + (y + 4)^2 = \frac{2025}{169}$

This is the equation of the required circle.

The equation of the circle circumscribing the triangle whose sides are the lines $y = x + 2$, $3y = 4x$, $2y = 3x$ is $\mathbf{x^2 + y^2 - 46x + 22y = 0}$ .

Solution:

The vertices of the triangle are the points of intersection of the given lines.

Let the lines be:

L1: $y = x + 2$

L2: $3y = 4x \implies y = \frac{4}{3}x$

L3: $2y = 3x \implies y = \frac{3}{2}x$

Find the intersection of L1 and L2:

Substitute $y = x + 2$ into $y = \frac{4}{3}x$:

$x + 2 = \frac{4}{3}x$

$3(x + 2) = 4x$

$3x + 6 = 4x$

$6 = x$

Substitute $x = 6$ into $y = x + 2$:

$y = 6 + 2 = 8$

Vertex A is $(6, 8)$.

Find the intersection of L1 and L3:

Substitute $y = x + 2$ into $y = \frac{3}{2}x$:

$x + 2 = \frac{3}{2}x$

$2(x + 2) = 3x$

$2x + 4 = 3x$

$4 = x$

Substitute $x = 4$ into $y = x + 2$:

$y = 4 + 2 = 6$

Vertex B is $(4, 6)$.

Find the intersection of L2 and L3:

Substitute $y = \frac{4}{3}x$ into $y = \frac{3}{2}x$:

$\frac{4}{3}x = \frac{3}{2}x$

$\frac{4}{3}x - \frac{3}{2}x = 0$

$\left(\frac{8 - 9}{6}\right)x = 0$

$-\frac{1}{6}x = 0 \implies x = 0$

Substitute $x = 0$ into $y = \frac{4}{3}x$:

$y = \frac{4}{3}(0) = 0$

Vertex C is $(0, 0)$.

The vertices of the triangle are $(6, 8)$, $(4, 6)$, and $(0, 0)$.

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

Since the circle passes through the vertex $(0, 0)$, substitute $x=0, y=0$ into the equation:

$0^2 + 0^2 + 2g(0) + 2f(0) + c = 0 \implies c = 0$.

The equation becomes $x^2 + y^2 + 2gx + 2fy = 0$.

Since the circle passes through the vertex $(4, 6)$, substitute $x=4, y=6$ into the equation:

$4^2 + 6^2 + 2g(4) + 2f(6) = 0$

$16 + 36 + 8g + 12f = 0$

$52 + 8g + 12f = 0$

Divide by 4:

Since the circle passes through the vertex $(6, 8)$, substitute $x=6, y=8$ into the equation:

$6^2 + 8^2 + 2g(6) + 2f(8) = 0$

$36 + 64 + 12g + 16f = 0$

$100 + 12g + 16f = 0$

Divide by 4:

Now solve the system of linear equations (i) and (ii) for $g$ and $f$.

Multiply equation (i) by 3 and equation (ii) by 2:

$3 \times (2g + 3f = -13) \implies 6g + 9f = -39$

$2 \times (3g + 4f = -25) \implies 6g + 8f = -50$

Subtract the second new equation from the first new equation:

$(6g + 9f) - (6g + 8f) = -39 - (-50)$

$f = -39 + 50$

$f = 11$

Substitute $f = 11$ into equation (i):

$2g + 3(11) = -13$

$2g + 33 = -13$

$2g = -13 - 33$

$2g = -46$

$g = -23$

Substitute the values of $g = -23$, $f = 11$, and $c = 0$ into the general equation of the circle $x^2 + y^2 + 2gx + 2fy + c = 0$:

$x^2 + y^2 + 2(-23)x + 2(11)y + 0 = 0$

$x^2 + y^2 - 46x + 22y = 0$

This is the equation of the circle circumscribing the given triangle.

The length of the string and distance between the pins are 6 cm and $2\sqrt{5}$ cm respectively.

Solution:

In the string method of drawing an ellipse, the two pins are located at the foci of the ellipse.

The endless string is passed over these two pins, and a pencil traces the ellipse while keeping the string taut. The length of the string is constant and is equal to the sum of the distances from any point on the ellipse to the two foci.

By the definition of an ellipse, the sum of the distances from any point on the ellipse to the two foci is equal to the length of the major axis, which is denoted by $2a$.

The given lengths of the axes are 6 cm and 4 cm.

The major axis is the longer axis, so the length of the major axis is $2a = 6$ cm.

The minor axis is the shorter axis, so the length of the minor axis is $2b = 4$ cm.

Thus, the length of the string is equal to the length of the major axis.

The distance between the pins is the distance between the two foci. This distance is denoted by $2c$, where $c$ is the distance from the center to each focus.

We have $2a = 6 \implies a = 3$ cm.

We have $2b = 4 \implies b = 2$ cm.

For an ellipse, the relationship between $a$, $b$, and $c$ is $a^2 = b^2 + c^2$.

We need to find $c$ to determine the distance between the foci ($2c$).

$3^2 = 2^2 + c^2$

$9 = 4 + c^2$

$c^2 = 9 - 4$

$c^2 = 5$

$c = \sqrt{5}$ cm (Since $c > 0$)

The distance between the foci (the pins) is $2c$.

From (i) and (ii), the length of the string is 6 cm and the distance between the pins is $2\sqrt{5}$ cm.

The equation of the ellipse having foci $(0, 1)$, $(0, –1)$ and minor axis of length 1 is $\mathbf{20x^2 + 4y^2 = 5}$ .

Solution:

The foci of the ellipse are given as $(0, 1)$ and $(0, -1)$.

Since the foci lie on the y-axis, the major axis of the ellipse is along the y-axis.

The center of the ellipse is the midpoint of the segment joining the foci.

Center $= \left(\frac{0+0}{2}, \frac{1+(-1)}{2}\right) = (0, 0)$.

The foci are at $(0, \pm c)$. Comparing with the given foci, we have $c = 1$.

Thus, $c^2 = 1^2 = 1$.

The length of the minor axis is given as 1. The length of the minor axis is $2b$.

So, $b = \frac{1}{2}$.

Thus, $b^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4}$.

For an ellipse with the major axis along the y-axis and center at the origin, the relationship between $a$, $b$, and $c$ is $a^2 = b^2 + c^2$.

Substitute the values of $b^2$ and $c^2$:

$a^2 = \frac{1 + 4}{4} = \frac{5}{4}$.

Since the major axis is along the y-axis and the center is at $(0, 0)$, the standard equation of the ellipse is $\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$.

Substitute the values of $a^2 = \frac{5}{4}$ and $b^2 = \frac{1}{4}$ into the standard equation:

$\frac{x^2}{1/4} + \frac{y^2}{5/4} = 1$

$4x^2 + \frac{4y^2}{5} = 1$

Multiply the entire equation by 5 to clear the fraction:

$5(4x^2) + 5\left(\frac{4y^2}{5}\right) = 5(1)$

$20x^2 + 4y^2 = 5$

This is the equation of the required ellipse.

The equation of the parabola having focus at $(-1, -2)$ and the directrix $x – 2y + 3 = 0$ is $\mathbf{4x^2 + 4xy + y^2 + 4x + 32y + 16 = 0}$ .

Solution:

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

The focus is $F(-1, -2)$.

The directrix is the line $L: x - 2y + 3 = 0$.

By the definition of a parabola, the distance from any point on the parabola to the focus is equal to the distance from that point to the directrix.

Distance from $P(x, y)$ to the focus $F(-1, -2)$ (PF):

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

Distance from $P(x, y)$ to the directrix $x - 2y + 3 = 0$ (PL):

$PL = \frac{|x - 2y + 3|}{\sqrt{1^2 + (-2)^2}} = \frac{|x - 2y + 3|}{\sqrt{1 + 4}} = \frac{|x - 2y + 3|}{\sqrt{5}}$

Equating the distances $PF = PL$:

Square both sides of equation (i):

$(\sqrt{(x + 1)^2 + (y + 2)^2})^2 = \left(\frac{|x - 2y + 3|}{\sqrt{5}}\right)^2$

$(x + 1)^2 + (y + 2)^2 = \frac{(x - 2y + 3)^2}{5}$

Expand the squared terms:

$(x^2 + 2x + 1) + (y^2 + 4y + 4) = \frac{x^2 + (-2y)^2 + 3^2 + 2(x)(-2y) + 2(x)(3) + 2(-2y)(3)}{5}$

$x^2 + y^2 + 2x + 4y + 5 = \frac{x^2 + 4y^2 + 9 - 4xy + 6x - 12y}{5}$

Multiply both sides by 5:

$5(x^2 + y^2 + 2x + 4y + 5) = x^2 + 4y^2 + 9 - 4xy + 6x - 12y$

$5x^2 + 5y^2 + 10x + 20y + 25 = x^2 + 4y^2 - 4xy + 6x - 12y + 9$

Move all terms to the left side:

$5x^2 - x^2 + 5y^2 - 4y^2 + 10x - 6x + 20y + 12y + 25 - 9 + 4xy = 0$

$4x^2 + y^2 + 4x + 32y + 16 + 4xy = 0$

Rearrange the terms in a standard order:

$4x^2 + 4xy + y^2 + 4x + 32y + 16 = 0$

This is the equation of the parabola.

The equation of the hyperbola with vertices at $(0, \pm 6)$ and eccentricity $\frac{5}{3}$ is $\mathbf{\frac{y^2}{36} - \frac{x^2}{64} = 1}$ and its foci are $\mathbf{(0, \pm 10)}$ .

Solution:

The given vertices of the hyperbola are $(0, \pm 6)$.

Since the vertices are on the y-axis and are symmetric about the origin, the center of the hyperbola is at $(0, 0)$ and the transverse axis is along the y-axis.

The standard equation of such a hyperbola is $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$.

The vertices are $(0, \pm a)$. Comparing with $(0, \pm 6)$, we get:

Thus, $a^2 = 6^2 = 36$.

The eccentricity is given as $e = \frac{5}{3}$.

For a hyperbola, the eccentricity is defined as $e = \frac{c}{a}$, where $c$ is the distance from the center to each focus.

Substitute the given value of $e$ and the value of $a$ into equation (i):

$\frac{5}{3} = \frac{c}{6}$

Solve for $c$:

$c = 6 \times \frac{5}{3} = 2 \times 5 = 10$

The foci of the hyperbola are at $(0, \pm c)$ because the transverse axis is along the y-axis and the center is at $(0, 0)$.

So, the foci are at $(0, \pm 10)$, which are the points $(0, 10)$ and $(0, -10)$.

For a hyperbola, the relationship between $a$, $b$, and $c$ is $c^2 = a^2 + b^2$.

We need to find $b^2$ to write the equation of the hyperbola. We have $a^2 = 36$ and $c = 10$ (so $c^2 = 100$).

$100 = 36 + b^2$

$b^2 = 100 - 36$

$b^2 = 64$

Substitute the values of $a^2 = 36$ and $b^2 = 64$ into the standard equation of the hyperbola $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$:

$\frac{y^2}{36} - \frac{x^2}{64} = 1$

This is the equation of the hyperbola.

The correct option is (C).

Solution:

The center of the circle is given as $(h, k) = (1, 2)$.

The circle passes through the point $(x, y) = (4, 6)$.

The radius ($r$) of the circle is the distance between the center $(1, 2)$ and the point $(4, 6)$.

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

$r = \sqrt{(4 - 1)^2 + (6 - 2)^2}$

$r = \sqrt{(3)^2 + (4)^2}$

$r = \sqrt{9 + 16}$

$r = \sqrt{25}$

$r = 5$

The radius of the circle is 5 units.

The area of the circle is given by the formula $A = \pi r^2$.

$A = \pi (5)^2$

$A = \pi (25)$

$A = 25\pi$

The area of the circle is $25\pi$ square units.

Comparing this with the given options, we find that option (C) matches the calculated area.

The correct option is (C).

Solution:

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

Since the circle touches both the x-axis and the y-axis, the distance from the center to the x-axis is $|k|$ and the distance from the center to the y-axis is $|h|$. Both of these distances must be equal to the radius $r$.

So, $|h| = r$ and $|k| = r$. This means the center is at $(\pm r, \pm r)$.

The circle passes through the point $(3, 6)$, which is in the first quadrant ($x > 0$, $y > 0$). For the circle to pass through a point in the first quadrant and touch both axes, its center must also be in the first quadrant.

Thus, the center of the circle is $(r, r)$ and the radius is $r$.

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

Since the circle passes through the point $(3, 6)$, substitute $x = 3$ and $y = 6$ into the equation:

Expand the terms in equation (i):

$(9 - 6r + r^2) + (36 - 12r + r^2) = r^2$

$9 - 6r + r^2 + 36 - 12r + r^2 = r^2$

Combine like terms:

$2r^2 - 18r + 45 = r^2$

Subtract $r^2$ from both sides to form a quadratic equation in $r$:

$r^2 - 18r + 45 = 0$

We can solve this quadratic equation for $r$ by factoring. We look for two numbers that multiply to 45 and add up to -18. These numbers are -3 and -15.

So, the equation can be factored as:

$(r - 3)(r - 15) = 0$

This gives two possible values for the radius $r$:

$r - 3 = 0 \implies r = 3$

or

$r - 15 = 0 \implies r = 15$

Case 1: $r = 3$.

The center is $(3, 3)$ and the radius is 3. The equation of the circle is:

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

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

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

Case 2: $r = 15$.

The center is $(15, 15)$ and the radius is 15. The equation of the circle is:

$(x - 15)^2 + (y - 15)^2 = 15^2$

$x^2 - 30x + 225 + y^2 - 30y + 225 = 225$

$x^2 + y^2 - 30x - 30y + 225 = 0$

Comparing the calculated equations with the given options:

• Option (A) is $x^2 + y^2 + 6x + 6y + 3 = 0$.
• Option (B) is $x^2 + y^2 – 6x – 6y – 9 = 0$.
• Option (C) is $x^2 + y^2 – 6x – 6y + 9 = 0$.
• Option (D) is none of these.

We see that the equation $x^2 + y^2 – 6x – 6y + 9 = 0$ from Case 1 matches option (C).

Let's verify that the point $(3, 6)$ lies on the circle from option (C):

$3^2 + 6^2 - 6(3) - 6(6) + 9 = 9 + 36 - 18 - 36 + 9 = 45 - 54 + 9 = 0$.

Since the equation evaluates to 0, the point $(3, 6)$ lies on the circle $x^2 + y^2 – 6x – 6y + 9 = 0$. This circle has center $(3, 3)$ and radius 3, so it touches both axes. Thus, this is a valid solution.

Option (C) is the correct answer.

Given:

Center of the circle is on the y-axis.

The circle passes through the points $(0, 0)$ and $(2, 3)$.

To Find:

The equation of the circle.

Solution:

Since the center of the circle lies on the y-axis, its x-coordinate is 0. Let the center be $(0, k)$.

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

Substituting the center $(0, k)$, the equation becomes:

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

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

The circle passes through the origin $(0, 0)$. Substituting this point into the equation:

$0^2 + (0 - k)^2 = r^2$

$k^2 = r^2$

The circle also passes through the point $(2, 3)$. Substituting this point into the equation:

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

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

Now we have two expressions for $r^2$: $r^2 = k^2$ and $r^2 = 4 + (3 - k)^2$. Equating these:

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

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

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

Subtract $k^2$ from both sides:

$0 = 13 - 6k$

$6k = 13$

$k = \frac{13}{6}$

Now we find $r^2$ using $r^2 = k^2$:

$r^2 = (\frac{13}{6})^2 = \frac{169}{36}$

Substitute the values of $k = \frac{13}{6}$ and $r^2 = \frac{169}{36}$ back into the circle equation $x^2 + (y - k)^2 = r^2$:

$x^2 + (y - \frac{13}{6})^2 = \frac{169}{36}$

Expand the term $(y - \frac{13}{6})^2$:

$(y - \frac{13}{6})^2 = y^2 - 2 \cdot y \cdot \frac{13}{6} + (\frac{13}{6})^2$

$(y - \frac{13}{6})^2 = y^2 - \frac{13}{3}y + \frac{169}{36}$

So the equation becomes:

$x^2 + y^2 - \frac{13}{3}y + \frac{169}{36} = \frac{169}{36}$

Subtract $\frac{169}{36}$ from both sides:

$x^2 + y^2 - \frac{13}{3}y = 0$

To remove the fraction, multiply the entire equation by 3:

$3(x^2 + y^2 - \frac{13}{3}y) = 3(0)$

$3x^2 + 3y^2 - 13y = 0$

Comparing our calculated equation $3x^2 + 3y^2 - 13y = 0$ with the given options:

(A) x² + y² + 13y = 0

(B) 3x² + 3y² + 13x + 3 = 0

(C) 6x² + 6y² – 13x = 0

(D) x² + y² + 13x + 3 = 0

None of the given options match the derived equation $3x^2 + 3y^2 - 13y = 0$.

Based on the calculation, there might be an error in the provided options.

The calculated equation is $3x^2 + 3y^2 - 13y = 0$.

Given:

Center of the circle is the origin $(0, 0)$.

The circle passes through the vertices of an equilateral triangle.

Length of the median of the equilateral triangle $= 3a$.

To Find:

The equation of the circle.

Solution:

For an equilateral triangle, the centroid, circumcenter, incenter, and orthocenter all coincide. In this case, the center of the circumscribing circle (which passes through the vertices) is the centroid.

The problem states that the center of the circle is the origin, which aligns with the fact that the centroid of the equilateral triangle is at the origin (the center of the circle).

The radius of the circumcircle of an equilateral triangle is $\frac{2}{3}$ times the length of its median (or altitude or height).

Let $r$ be the radius of the circle.

Using the given hint:

$r = \frac{2}{3} \times (\text{length of median})$

$r = \frac{2}{3} \times (3a)$

$r = 2a$

The equation of a circle with center $(h, k)$ and radius $r$ is $(x - h)^2 + (y - k)^2 = r^2$.

Here, the center is $(0, 0)$ and the radius is $r = 2a$.

Substituting these values into the equation:

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

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

Comparing this equation with the given options:

(A) $x^2 + y^2 = 9a^2$

(B) $x^2 + y^2 = 16a^2$

(C) $x^2 + y^2 = 4a^2$

(D) $x^2 + y^2 = a^2$

The calculated equation $x^2 + y^2 = 4a^2$ matches option (C).

The final answer is (C) $x^2 + y^2 = 4a^2$.

Given:

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

Directrix of the parabola is the line $y = 3$.

To Find:

The equation of the parabola.

Solution:

A parabola is defined as the locus of points that are equidistant from a fixed point (the focus) and a fixed line (the directrix).

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

The distance from $P(x, y)$ to 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 directrix is the line $y = 3$, which can be written as $y - 3 = 0$. The distance from a point $P(x, y)$ to the line $Ax + By + C = 0$ is $\frac{|Ax + By + C|}{\sqrt{A^2 + B^2}}$.

The distance from $P(x, y)$ to the directrix $y - 3 = 0$ is:

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

By the definition of the parabola, the distance from $P$ to the focus is equal to the distance from $P$ to the directrix:

$PF = PD$

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

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

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

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

Expand the squared terms on both sides:

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

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

Subtract $y^2$ and $9$ from both sides of the equation:

$x^2 + y^2 - y^2 + 6y - (-6y) + 9 - 9 = 0$

$x^2 + 6y + 6y = 0$

$x^2 + 12y = 0$

Rearrange the equation to match the options:

$x^2 = -12y$

Compare the derived equation with the given options:

(A) $x^2 = –12y$

(B) $x^2 = 12y$

(C) $y^2 = –12x$

(D) $y^2 = 12x$

The derived equation $x^2 = -12y$ matches option (A).

The final answer is (A) $x^2 = –12y$.

Given:

The equation of the parabola is $y^2 = 4ax$.

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

To Find:

The length of the latus rectum of the parabola.

Solution:

The equation of the parabola is given as $y^2 = 4ax$.

The length of the latus rectum for a parabola of the form $y^2 = 4ax$ is $4a$.

To find the length of the latus rectum, we need to determine the value of $a$.

Since the parabola passes through the point $(3, 2)$, the coordinates of this point must satisfy the equation of the parabola.

Substitute $x = 3$ and $y = 2$ into the equation $y^2 = 4ax$:

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

$4 = 12a$

Now, solve for $a$:

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

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

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

Substitute the value of $a = \frac{1}{3}$:

Length of latus rectum $= 4 \times \frac{1}{3} = \frac{4}{3}$

Comparing the calculated length with the given options:

(A) $\frac{2}{3}$

(B) $\frac{4}{3}$

(C) $\frac{1}{3}$

(D) 4

The calculated length $\frac{4}{3}$ matches option (B).

The final answer is (B) $\frac{4}{3}$.

Given:

Vertex of the parabola $V = (-3, 0)$.

Directrix of the parabola is the line $x + 5 = 0$, which can be written as $x = -5$.

To Find:

The equation of the parabola.

Solution:

The vertex of the parabola is $(h, k) = (-3, 0)$, so $h = -3$ and $k = 0$.

The directrix is a vertical line $x = -5$. This indicates that the axis of symmetry of the parabola is horizontal, and the parabola opens either to the right or to the left.

The general form of the equation of a parabola with a horizontal axis of symmetry and vertex $(h, k)$ is $(y - k)^2 = 4p(x - h)$.

The equation of the directrix for this type of parabola is $x = h - p$.

We are given the vertex $(h, k) = (-3, 0)$ and the directrix $x = -5$.

Substitute the value of $h$ from the vertex into the directrix equation form:

$x = -3 - p$

We know the directrix is $x = -5$. So, we can equate the two expressions for $x$:

$-5 = -3 - p$

Solve for $p$:

$p = -3 - (-5)$

$p = -3 + 5$

$p = 2$

Since $p = 2 > 0$, the parabola opens to the right.

Now, substitute the vertex $(h, k) = (-3, 0)$ and $p = 2$ into the standard equation $(y - k)^2 = 4p(x - h)$:

$(y - 0)^2 = 4(2)(x - (-3))$

$y^2 = 8(x + 3)$

Comparing the derived equation with the given options:

(A) $y^2 = 8 (x + 3)$

(B) $x^2 = 8 (y + 3)$

(C) $y^2 = – 8 (x + 3)$

(D) $y^2 = 8 ( x + 5)$

The derived equation $y^2 = 8(x + 3)$ matches option (A).

The final answer is (A) $y^2 = 8 (x + 3)$.

Given:

Focus $F = (1, -1)$.

Directrix is the line $x - y - 3 = 0$.

Eccentricity $e = \frac{1}{2}$.

To Find:

The equation of the ellipse.

Solution:

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

By the definition of a conic section, the distance from $P$ to the focus ($PF$) is $e$ times the distance from $P$ to the directrix ($PD$).

$PF = e \cdot PD$

The distance from $P(x, y)$ to the focus $F(1, -1)$ is:

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

The distance from $P(x, y)$ to the directrix $x - y - 3 = 0$ is:

$PD = \frac{|x - y - 3|}{\sqrt{1^2 + (-1)^2}} = \frac{|x - y - 3|}{\sqrt{1 + 1}} = \frac{|x - y - 3|}{\sqrt{2}}$

Using the relation $PF = e \cdot PD$:

$\sqrt{(x - 1)^2 + (y + 1)^2} = \frac{1}{2} \cdot \frac{|x - y - 3|}{\sqrt{2}}$

$\sqrt{(x - 1)^2 + (y + 1)^2} = \frac{|x - y - 3|}{2\sqrt{2}}$

Square both sides:

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

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

$x^2 + y^2 - 2x + 2y + 2 = \frac{x^2 + y^2 + (-3)^2 + 2(x)(-y) + 2(-y)(-3) + 2(-3)(x)}{8}$

$x^2 + y^2 - 2x + 2y + 2 = \frac{x^2 + y^2 + 9 - 2xy + 6y - 6x}{8}$

Multiply both sides by 8:

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

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

Move all terms to the left side:

$8x^2 - x^2 + 8y^2 - y^2 - (-2xy) - 16x - (-6x) + 16y - 6y + 16 - 9 = 0$

$7x^2 + 7y^2 + 2xy - 16x + 6x + 10y + 7 = 0$

$7x^2 + 7y^2 + 2xy - 10x + 10y + 7 = 0$

Comparing the derived equation with the given options, it matches option (A).

The final answer is (A) $7x^2 + 2xy + 7y^2 – 10x + 10y + 7 = 0$.

Given:

The equation of the ellipse is $3x^2 + y^2 = 12$.

To Find:

The length of the latus rectum of the ellipse.

Solution:

First, we need to convert the given equation of the ellipse into the standard form, which is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ or $\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$ (where $a$ is the semi-major axis and $b$ is the semi-minor axis, and $a > b$).

Divide the given equation by 12:

$\frac{3x^2}{12} + \frac{y^2}{12} = \frac{12}{12}$

$\frac{x^2}{4} + \frac{y^2}{12} = 1$

Comparing this equation with the standard form $\frac{x^2}{B} + \frac{y^2}{A} = 1$, where the major axis is along the y-axis (since the denominator under $y^2$ is larger), we have:

$a^2 = 12$

$b^2 = 4$

From these values, we find $a$ and $b$:

$a = \sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}$

$b = \sqrt{4} = 2$

Since $a = 2\sqrt{3}$ and $b = 2$, we have $a > b$, which is consistent for the semi-major and semi-minor axes.

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

Substitute the values of $a$ and $b^2$ into the formula:

Length of latus rectum $= \frac{2 \times 4}{2\sqrt{3}}$

Length of latus rectum $= \frac{8}{2\sqrt{3}}$

Length of latus rectum $= \frac{4}{\sqrt{3}}$

Comparing the calculated length with the given options:

(A) 4

(B) 3

(C) 8

(D) $\frac{4}{\sqrt{3}}$

The calculated length $\frac{4}{\sqrt{3}}$ matches option (D).

The final answer is (D) $\frac{4}{\sqrt{3}}$.

Given:

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

The condition $a < b$ is given, which means $b^2 > a^2$.

$e$ is the eccentricity of the ellipse.

To Find:

The relationship between $a$, $b$, and $e$ for the given ellipse.

Solution:

The standard equation of an ellipse is $\frac{x^2}{A^2} + \frac{y^2}{B^2} = 1$, where the major axis is along the axis corresponding to the larger denominator.

In this problem, the equation is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, and we are given $a < b$.

Since $a < b$, it follows that $a^2 < b^2$. This means the denominator under $y^2$ is larger than the denominator under $x^2$.

Therefore, the major axis of the ellipse is along the y-axis.

For an ellipse with the major axis along the y-axis (i.e., $b$ is the semi-major axis and $a$ is the semi-minor axis), the relationship between $a$, $b$, and the eccentricity $e$ is given by:

$a^2 = b^2(1 - e^2)$

Comparing this derived relationship with the given options:

(A) $b^2 = a^2 (1 – e^2)$ (This is for the case where the major axis is along the x-axis, $a > b$)

(B) $a^2 = b^2 (1 – e^2)$ (This matches the case where the major axis is along the y-axis, $b > a$)

(C) $a^2 = b^2 (e^2 – 1)$ (This form is related to a hyperbola)

(D) $b^2 = a^2 (e^2 – 1)$ (This form is related to a hyperbola)

The derived relationship $a^2 = b^2 (1 – e^2)$ matches option (B).

The final answer is (B) $a^2 = b^2 (1 – e^2)$.

Given:

Length of the latus rectum of the hyperbola $= 8$.

Length of the conjugate axis $= \frac{1}{2} \times (\text{Distance between the foci})$.

To Find:

The eccentricity $e$ of the hyperbola.

Solution:

Let the equation of the hyperbola be $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ (or $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$). In either case, $2a$ is the length of the transverse axis and $2b$ is the length of the conjugate axis. The distance between the foci is $2ae$, and the length of the latus rectum is $\frac{2b^2}{a}$.

From the given information about the latus rectum:

Length of latus rectum $= \frac{2b^2}{a} = 8$

$2b^2 = 8a$

$b^2 = 4a$ ... (i)

From the given information about the conjugate axis and the distance between foci:

Length of conjugate axis $= 2b$

Distance between foci $= 2ae$

According to the problem:

$2b = \frac{1}{2} (2ae)$

$2b = ae$

$b = \frac{ae}{2}$ ... (ii)

Substitute the expression for $b$ from equation (ii) into equation (i):

$(\frac{ae}{2})^2 = 4a$

$\frac{a^2e^2}{4} = 4a$

Since $a$ is the semi-transverse axis length of a hyperbola, $a \neq 0$. We can divide both sides by $a$:

$\frac{ae^2}{4} = 4$

$ae^2 = 16$

$a = \frac{16}{e^2}$ ... (iii)

Now, substitute the expression for $a$ from equation (iii) into equation (ii):

$b = \frac{(\frac{16}{e^2})e}{2}$

$b = \frac{\frac{16}{e}}{2}$

$b = \frac{16}{2e}$

$b = \frac{8}{e}$ ... (iv)

The relationship between $a$, $b$, and the eccentricity $e$ for a hyperbola is $b^2 = a^2(e^2 - 1)$.

Substitute the expressions for $a$ and $b$ from equations (iii) and (iv) into this relationship:

$(\frac{8}{e})^2 = (\frac{16}{e^2})^2 (e^2 - 1)$

$\frac{64}{e^2} = \frac{256}{e^4} (e^2 - 1)$

Since $e \neq 0$, multiply both sides by $e^4$:

$64e^2 = 256(e^2 - 1)$

Divide both sides by 64:

$\frac{64e^2}{64} = \frac{256(e^2 - 1)}{64}$

$e^2 = 4(e^2 - 1)$

$e^2 = 4e^2 - 4$

$4e^2 - e^2 = 4$

$3e^2 = 4$

$e^2 = \frac{4}{3}$

Take the square root of both sides. Since eccentricity $e > 1$ for a hyperbola, we take the positive root:

$e = \sqrt{\frac{4}{3}} = \frac{\sqrt{4}}{\sqrt{3}} = \frac{2}{\sqrt{3}}$

Comparing the calculated eccentricity with the given options:

(A) $\frac{4}{3}$

(B) $\frac{4}{\sqrt{3}}$

(C) $\frac{2}{\sqrt{3}}$

(D) none of these

The calculated eccentricity $\frac{2}{\sqrt{3}}$ matches option (C).

The final answer is (C) $\frac{2}{\sqrt{3}}$.

Given:

Distance between the foci of the hyperbola $= 16$.

Eccentricity of the hyperbola $e = \sqrt{2}$.

To Find:

The equation of the hyperbola.

Solution:

The distance between the foci of a hyperbola is given by $2ae$, where $a$ is the length of the semi-transverse axis and $e$ is the eccentricity.

We are given that the distance between foci is 16, so:

$2ae = 16$

We are given the eccentricity $e = \sqrt{2}$. Substitute this value into the equation:

$2a(\sqrt{2}) = 16$

$2\sqrt{2}a = 16$

Solve for $a$:

$a = \frac{16}{2\sqrt{2}} = \frac{8}{\sqrt{2}}$

Rationalize the denominator:

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

So, $a = 4\sqrt{2}$.

Now find $a^2$:

$a^2 = (4\sqrt{2})^2 = 4^2 \times (\sqrt{2})^2 = 16 \times 2 = 32$

For a hyperbola, the relationship between $a$, $b$ (semi-conjugate axis), and $e$ is $b^2 = a^2(e^2 - 1)$.

We have $a^2 = 32$ and $e = \sqrt{2}$. Substitute these values:

$b^2 = 32((\sqrt{2})^2 - 1)$

$b^2 = 32(2 - 1)$

$b^2 = 32(1)$

$b^2 = 32$

Since the foci are typically assumed to be on the x-axis unless stated otherwise, the standard equation of the hyperbola is usually $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.

Substitute the values $a^2 = 32$ and $b^2 = 32$ into the standard equation:

$\frac{x^2}{32} - \frac{y^2}{32} = 1$

Multiply the entire equation by 32:

$32 \left( \frac{x^2}{32} - \frac{y^2}{32} \right) = 32 \times 1$

$x^2 - y^2 = 32$

Comparing the derived equation with the given options:

(A) $x^2 – y^2 = 32$

(B) $\frac{x^2}{4} - \frac{y^2}{9} = 1$

(C) $2x – 3y^2 = 7$ (This is not the equation of a hyperbola)

(D) none of these

The derived equation $x^2 - y^2 = 32$ matches option (A).

Note that since $a^2 = b^2$, this hyperbola is a rectangular hyperbola.

The final answer is (A) $x^2 – y^2 = 32$.

Given:

Eccentricity $e = \frac{3}{2}$.

Foci are at $(\pm 2, 0)$.

To Find:

The equation of the hyperbola.

Solution:

The foci of the hyperbola are given as $(\pm 2, 0)$. This indicates that the center of the hyperbola is at the origin $(0, 0)$, and the transverse axis is along the x-axis.

The distance from the center to each focus is $c$. Thus, $c = 2$.

For a hyperbola with foci on the x-axis centered at the origin, the distance from the center to the focus is also given by $ae$, where $a$ is the length of the semi-transverse axis and $e$ is the eccentricity.

Substitute the given values $e = \frac{3}{2}$ and $c = 2$ into equation (i):

$a \left(\frac{3}{2}\right) = 2$

Solve for $a$:

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

Calculate $a^2$:

$a^2 = \left(\frac{4}{3}\right)^2 = \frac{16}{9}$

For a hyperbola, the relationship between the semi-transverse axis $a$, the semi-conjugate axis $b$, and the eccentricity $e$ is $b^2 = a^2(e^2 - 1)$.

Substitute the values of $a^2 = \frac{16}{9}$ and $e = \frac{3}{2}$ into this relationship:

$b^2 = \frac{16}{9} \left( \left(\frac{3}{2}\right)^2 - 1 \right)$

$b^2 = \frac{16}{9} \left( \frac{9}{4} - 1 \right)$

$b^2 = \frac{16}{9} \left( \frac{9 - 4}{4} \right)$

$b^2 = \frac{16}{9} \left( \frac{5}{4} \right)$

$b^2 = \frac{\cancel{16}^{4} \times 5}{9 \times \cancel{4}^{1}}$

$b^2 = \frac{4 \times 5}{9} = \frac{20}{9}$

The standard equation of a hyperbola with center at the origin and transverse axis along the x-axis is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.

Substitute the calculated values $a^2 = \frac{16}{9}$ and $b^2 = \frac{20}{9}$ into the standard equation:

$\frac{x^2}{\frac{16}{9}} - \frac{y^2}{\frac{20}{9}} = 1$

$\frac{9x^2}{16} - \frac{9y^2}{20} = 1$

Now let's examine option (A) and see if it matches our derived equation:

(A) $\frac{x^2}{4} - \frac{y^2}{5} = \frac{4}{9}$

To put this in the standard form $= 1$, divide both sides by $\frac{4}{9}$ (or multiply by $\frac{9}{4}$):

$\frac{x^2}{4} \times \frac{9}{4} - \frac{y^2}{5} \times \frac{9}{4} = \frac{4}{9} \times \frac{9}{4}$

$\frac{9x^2}{16} - \frac{9y^2}{20} = 1$

This matches our derived equation.

Comparing our result with the given options, option (A) is equivalent to the equation we found.

The final answer is (A) $\frac{x^2}{4} - \frac{y^2}{5} = \frac{4}{9}$.