Ex 10.1 - Question 1

Question: Find the equation of the circle with centre $(0, 2)$ and radius $2$.

Solution:

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

Given:

• Centre $(h, k) = (0, 2)$
• Radius $r = 2$

Substitute these values into the standard equation:

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

Ex 10.1 - Question 2

Question: Find the equation of the circle with centre $(-2, 3)$ and radius $4$.

Solution:

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

Given:

• Centre $(h, k) = (-2, 3)$
• Radius $r = 4$

Substitute the values into the formula:

Expanding the terms:

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

Ex 10.1 - Question 3

Question: Find the equation of the circle with centre $\left(\frac{1}{2}, \frac{1}{4}\right)$ and radius $\frac{1}{12}$.

Solution:

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

Given:

• Centre $(h, k) = \left(\frac{1}{2}, \frac{1}{4}\right)$
• Radius $r = \frac{1}{12}$

Substitute the values into the formula:

Rearranging and grouping terms:

Subtract $\frac{1}{144}$ from both sides:

Finding a common denominator ($144$) for the constant terms:

So the equation becomes:

Multiply the entire equation by $36$ to clear fractions:

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

Ex 10.1 - Question 4

Question: Find the equation of the circle with centre $(1, 1)$ and radius $\sqrt{2}$.

Solution:

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

Given:

• Centre $(h, k) = (1, 1)$
• Radius $r = \sqrt{2}$

Substituting these values into the standard equation:

Expanding the terms:

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

Ex 10.1 - Question 5

Question: Find the equation of the circle with centre $(-a, -b)$ and radius $\sqrt{a^2 - b^2}$.

Solution:

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

Given:

• Centre $(h, k) = (-a, -b)$
• Radius $r = \sqrt{a^2 - b^2}$

Substituting these values into the standard equation:

Expanding the binomials:

Subtracting $a^2$ from both sides and rearranging:

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

Ex 10.1 - Question 6

Question: Find the centre and radius of the circle: $(x + 5)^2 + (y - 3)^2 = 36$

Solution:

The given equation is:

Compare this equation with the standard equation of a circle, which is:

Rewriting the given equation to match the standard form:

By comparison, we get:

• Centre $(h, k) = (-5, 3)$
• Radius $r = 6$

Answer: Centre is $(-5, 3)$ and radius is $6$.

Ex 10.1 - Question 7

Question: Find the centre and radius of the circle: $x^2 + y^2 - 4x - 8y - 45 = 0$

Solution:

The given equation of the circle is:

Rearranging the terms by grouping $x$ terms and $y$ terms together:

Completing the square for both $x$ and $y$ expressions:

• For $x$: add and subtract $\left(\frac{-4}{2}\right)^2 = (-2)^2 = 4$
• For $y$: add and subtract $\left(\frac{-8}{2}\right)^2 = (-4)^2 = 16$

Adding these values to both sides of the equation:

Comparing this with the standard equation of a circle $(x - h)^2 + (y - k)^2 = r^2$:

• Centre $(h, k) = (2, 4)$
• Radius $r^2 = 65 \implies r = \sqrt{65}$

Answer: Centre is $(2, 4)$ and radius is $\sqrt{65}$.

Ex 10.1 - Question 8

Question: Find the centre and radius of the circle: $x^2 + y^2 - 8x + 10y - 12 = 0$

Solution:

Given the equation of the circle:

Group the $x$ terms and $y$ terms together:

Complete the square for both $x$ and $y$:

Comparing this with the standard equation of a circle $(x - h)^2 + (y - k)^2 = r^2$, we get:

• Centre $(h, k) = (4, -5)$
• Radius $r = \sqrt{53}$

Ex 10.1 - Question 9

Question: Find the centre and radius of the circle: $2x^2 + 2y^2 - x = 0$

Solution:

Given the equation of the circle:

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

Rearrange and group the $x$ terms:

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

Comparing with the standard form $(x - h)^2 + (y - k)^2 = r^2$:

• Centre $(h, k) = \left(\frac{1}{4}, 0\right)$
• Radius $r = \frac{1}{4}$

Ex 10.1 - Question 10

Question: Find the equation of the circle passing through the points $(4, 1)$ and $(6, 5)$ and whose centre is on the line $4x + y = 16$.

Solution:

Let the equation of the circle be:

Step 1: Since the centre $(h, k)$ lies on the line $4x + y = 16$:

Step 2: Since the circle passes through $(4, 1)$ and $(6, 5)$, these points satisfy the circle's equation:

Step 3: Equating Equation 2 and Equation 3 (since both equal $r^2$):

Step 4: Substitute $k = 16 - 4h$ into Equation 4:

Now find $k$:

Step 5: Find the radius $r^2$ using Equation 2:

Step 6: Write the final equation of the circle:

Ex 10.1 - Question 11

Question: Find the equation of the circle passing through the points $(2, 3)$ and $(-1, 1)$ and whose centre is on the line $x - 3y - 11 = 0$.

Solution:

Let the equation of the circle be:

Step 1: Since the centre $(h, k)$ lies on the line $x - 3y - 11 = 0$:

Step 2: Since the circle passes through $(2, 3)$ and $(-1, 1)$, substitute these points into the circle equation:

Step 3: Equating Equation 2 and Equation 3:

Step 4: Substitute $h = 3k + 11$ into Equation 4:

Now find $h$:

Step 5: Find the radius $r^2$ using Equation 3:

Step 6: Write the final equation of the circle:

Ex 10.1 - Question 12

Question: Find the equation of the circle with radius $5$ whose centre lies on $x$-axis and passes through the point $(2, 3)$.

Solution:

Since the centre of the circle lies on the $x$-axis, let the coordinates of the centre be $(h, 0)$.

Given that the radius of the circle is $r = 5$ and it passes through the point $(2, 3)$, the distance from the centre $(h, 0)$ to the point $(2, 3)$ must equal the radius $5$:

$$\sqrt{(2 - h)^2 + (3 - 0)^2} = 5$$

Squaring both sides:

$$(2 - h)^2 + 3^2 = 5^2$$

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

$$(2 - h)^2 = 16$$

Taking the square root on both sides:

$$2 - h = \pm 4$$

This gives two possible values for $h$:

1. If $2 - h = 4 \implies h = -2$, so the centre is $(-2, 0)$.
2. If $2 - h = -4 \implies h = 6$, so the centre is $(6, 0)$.

Case 1: Centre is $(-2, 0)$ and radius $r = 5$.

The equation of the circle is:

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

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

Case 2: Centre is $(6, 0)$ and radius $r = 5$.

The equation of the circle is:

$$(x - 6)^2 + (y - 0)^2 = 5^2$$

$x^2 - 12x + 36 + y^2 = 25 \implies x^2 + y^2 - 12x + 11 = 0$

Thus, the equations of the circles are $x^2 + y^2 + 4x - 21 = 0$ and $x^2 + y^2 - 12x + 11 = 0$.

Ex 10.1 - Question 13

Question: Find the equation of the circle passing through $(0, 0)$ and making intercepts $a$ and $b$ on the coordinate axes.

Solution:

Let the general equation of a circle be:

$$x^2 + y^2 + 2gx + 2fy + c = 0$$

Since the circle passes through the origin $(0, 0)$:

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

The circle makes intercepts $a$ and $b$ on the $x$ and $y$ axes respectively. The length of the intercept made on the $x$-axis is given by $2\sqrt{g^2 - c}$ (or simply, the points of intersection with the $x$-axis are $(a, 0)$ and $(0,0)$ since it passes through the origin).

Thus, substituting the point $(a, 0)$ into the circle's equation ($c=0$):

$$a^2 + 0^2 + 2g(a) + 2f(0) + 0 = 0 \implies a^2 + 2ga = 0$$

Since $a \neq 0$ (assuming non-zero intercept), we get $2g = -a$.

Similarly, substituting the point $(0, b)$ into the equation:

$$0^2 + b^2 + 2g(0) + 2f(b) + 0 = 0 \implies b^2 + 2fb = 0$$

Since $b \neq 0$, we get $2f = -b$.

Substituting $2g = -a$, $2f = -b$, and $c = 0$ back into the general circle equation:

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

Ex 10.1 - Question 14

Question: Find the equation of a circle with centre $(2, 2)$ and passes through the point $(4, 5)$.

Solution:

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

The radius $r$ of the circle is the distance between the centre $(2, 2)$ and the point $(4, 5)$ lying on the circumference:

$$r = \sqrt{(4 - 2)^2 + (5 - 2)^2} = \sqrt{2^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13}$$

Using the standard equation of a circle $(x - h)^2 + (y - k)^2 = r^2$:

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

Expanding the terms:

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

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

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

Thus, the equation of the circle is $x^2 + y^2 - 4x - 4y = 5$.

Ex 10.1 - Question 15

Question: Does the point $(-2.5, 3.5)$ lie inside, outside or on the circle $x^2 + y^2 = 25$?

Solution:

Given the circle equation:

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

The centre of this circle is $(0, 0)$ and its radius is $r = \sqrt{25} = 5$.

Let the given point be $P(-2.5, 3.5)$. We calculate the distance $d$ from the origin $(0, 0)$ to point $P$:

$$d = \sqrt{(-2.5 - 0)^2 + (3.5 - 0)^2} = \sqrt{(-2.5)^2 + (3.5)^2}$$

$$d = \sqrt{6.25 + 12.25} = \sqrt{18.5}$$

Comparing the distance $d$ with the radius $r$:

Since $\sqrt{18.5} < \sqrt{25}$ (i.e., $d < r$), the distance of the point from the centre is less than the radius of the circle.

Therefore, the point $(-2.5, 3.5)$ lies inside the circle.

Ex 10.2 - Question 1

Question: Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of the latus rectum for the equation: $y^2 = 12x$

Solution:

Given equation of the parabola is:

Comparing this equation with the standard equation of a right-handed parabola opening along the positive x-axis, $y^2 = 4ax$, we get:

Now, let us find the required parameters:

• Focus: The coordinates of the focus for $y^2 = 4ax$ are given by $(a, 0)$. Substituting $a = 3$, we get the focus at $(3, 0)$.
• Axis of the parabola: Since the equation contains $y^2$, the axis of symmetry is the $x\text{-axis}$.
• Equation of the directrix: The directrix is given by $x = -a$. Substituting $a = 3$, we get $x = -3$.
• Length of the latus rectum: The length of the latus rectum is given by $4a = 12$.

Answer Summary:
Focus: $(3, 0)$, Axis: $x\text{-axis}$, Directrix: $x = -3$, Length of Latus Rectum: $12$.

Ex 10.2 - Question 2

Question: Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of the latus rectum for the equation: $x^2 = 6y$

Solution:

Given equation of the parabola is:

Comparing this equation with the standard equation of an upward-opening parabola, $x^2 = 4ay$, we get:

Now, let us find the required parameters:

• Focus: The coordinates of the focus for $x^2 = 4ay$ are given by $(0, a)$. Substituting $a = \frac{3}{2}$, we get the focus at $\left(0, \frac{3}{2}\right)$.
• Axis of the parabola: Since the equation contains $x^2$, the axis of symmetry is the $y\text{-axis}$.
• Equation of the directrix: The directrix is given by $y = -a$. Substituting $a = \frac{3}{2}$, we get $y = -\frac{3}{2}$.
• Length of the latus rectum: The length of the latus rectum is given by $4a = 6$.

Answer Summary:
Focus: $\left(0, \frac{3}{2}\right)$, Axis: $y\text{-axis}$, Directrix: $y = -\frac{3}{2}$, Length of Latus Rectum: $6$.

Ex 10.2 - Question 3

Question: Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of the latus rectum for the equation: $y^2 = -8x$

Solution:

Given equation of the parabola is:

Comparing this equation with the standard equation of a left-handed parabola, $y^2 = -4ax$, we get:

Now, let us find the required parameters:

• Focus: The coordinates of the focus for $y^2 = -4ax$ are given by $(-a, 0)$. Substituting $a = 2$, we get the focus at $(-2, 0)$.
• Axis of the parabola: Since the equation contains $y^2$, the axis of symmetry is the $x\text{-axis}$.
• Equation of the directrix: The directrix is given by $x = a$. Substituting $a = 2$, we get $x = 2$.
• Length of the latus rectum: The length of the latus rectum is given by $4a = 8$.

Answer Summary:
Focus: $(-2, 0)$, Axis: $x\text{-axis}$, Directrix: $x = 2$, Length of Latus Rectum: $8$.

Ex 10.2 - Question 4

Question: Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of the latus rectum for the equation: $x^2 = -16y$

Solution:

Given equation of the parabola is:

Comparing this equation with the standard equation of a downward-opening parabola, $x^2 = -4ay$, we get:

Now, let us find the required parameters:

• Focus: The coordinates of the focus for $x^2 = -4ay$ are given by $(0, -a)$. Substituting $a = 4$, we get the focus at $(0, -4)$.
• Axis of the parabola: Since the equation contains $x^2$, the axis of symmetry is the $y\text{-axis}$.
• Equation of the directrix: The directrix is given by $y = a$. Substituting $a = 4$, we get $y = 4$.
• Length of the latus rectum: The length of the latus rectum is given by $4a = 16$.

Answer Summary:
Focus: $(0, -4)$, Axis: $y\text{-axis}$, Directrix: $y = 4$, Length of Latus Rectum: $16$.

Ex 10.2 - Question 5

Question: Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of the latus rectum for the equation: $y^2 = 10x$

Solution:

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

On comparing this equation with the standard equation of a parabola opening to the right, $y^2 = 4ax$, we get:

Now, let's find the required properties:

• Axis: The axis of the parabola is the $x$-axis.
• Focus: The coordinates of the focus are $(a, 0)$, which gives $\left(\frac{5}{2}, 0\right)$.
• Directrix: The equation of the directrix is $x = -a$, which gives $x = -\frac{5}{2}$.
• Length of the Latus rectum: The length is $4a = 10$.

Answer Summary:
Focus: $\left(\frac{5}{2}, 0\right)$, Axis: $x\text{-axis}$, Directrix: $x = -\frac{5}{2}$, Length of Latus Rectum: $10$.

Ex 10.2 - Question 6

Question: $x^2 = -9y$

Solution:

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

On comparing this equation with the standard equation of a parabola opening downwards, $x^2 = -4ay$, we get:

Now, let's find the required properties:

• Axis: The axis of the parabola is the $y$-axis.
• Focus: The coordinates of the focus are $(0, -a)$, which gives $\left(0, -\frac{9}{4}\right)$.
• Directrix: The equation of the directrix is $y = a$, which gives $y = \frac{9}{4}$.
• Length of the Latus rectum: The length is $4a = 9$.

Ex 10.2 - Question 7

Question: Focus $(6, 0)$; directrix $x = -6$

Solution:

Given that the focus is $(6, 0)$ and the directrix is $x = -6$.

Since the focus lies on the $x$-axis and the directrix is a vertical line parallel to the $y$-axis with a negative $x$-intercept, the parabola is symmetrical about the $x$-axis and opens to the right.

The standard equation of this type of parabola is:

From the focus coordinates $(a, 0) = (6, 0)$, we find $a = 6$.

Substituting $a = 6$ into the equation:

Ex 10.2 - Question 8

Question: Focus $(0, -3)$; directrix $y = 3$

Solution:

Given that the focus is $(0, -3)$ and the directrix is $y = 3$.

Since the focus lies on the $y$-axis with a negative coordinate and the directrix is a horizontal line with a positive $y$-intercept, the parabola opens downwards along the $y$-axis.

The standard equation of this type of parabola is:

From the focus coordinates $(0, -a) = (0, -3)$, we find $a = 3$.

Substituting $a = 3$ into the equation:

Ex 10.2 - Question 9

Question: Vertex $(0, 0)$; focus $(3, 0)$

Solution:

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

Since the focus lies on the positive $x$-axis, the parabola opens to the right. The standard equation of such a parabola is:

$$y^2 = 4ax$$

The coordinates of the focus are given as $(a, 0)$. By comparing, we get:

$$a = 3$$

Substituting $a = 3$ into the standard equation:

$$y^2 = 4(3)x \implies y^2 = 12x$$

Ex 10.2 - Question 10

Question: Vertex $(0, 0)$; focus $(-2, 0)$

Solution:

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

Since the focus lies on the negative $x$-axis, the parabola opens to the left. The standard equation of such a parabola is:

$$y^2 = -4ax$$

The coordinates of the focus are $(-a, 0)$. By comparing with $(-2, 0)$, we get:

$$a = 2$$

Substituting $a = 2$ into the standard equation:

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

Ex 10.2 - Question 11

Question: Vertex $(0, 0)$ passing through $(2, 3)$ and axis is along $x$-axis.

Solution:

Since the vertex is $(0, 0)$ and the axis is along the $x$-axis, the equation of the parabola is of the form:

$$y^2 = 4ax \quad \text{or} \quad y^2 = -4ax$$

Since the point $(2, 3)$ has a positive $x$-coordinate, the parabola opens to the right and its equation is:

$$y^2 = 4ax$$

Since the parabola passes through $(2, 3)$, we substitute $x = 2$ and $y = 3$ into the equation:

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

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

Substituting the value of $a$ back into the equation:

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

Ex 10.2 - Question 12

Question: Vertex $(0, 0)$, passing through $(5, 2)$ and symmetric with respect to $y$-axis.

Solution:

Since the vertex is $(0, 0)$ and it is symmetric with respect to the $y$-axis, the parabola opens upwards or downwards. The standard equation is:

$$x^2 = 4ay$$

Since the point $(5, 2)$ has a positive $y$-coordinate, the parabola opens upwards. Thus, we use:

$$x^2 = 4ay$$

Since the parabola passes through $(5, 2)$, we substitute $x = 5$ and $y = 2$ into the equation:

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

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

Substituting the value of $a$ back into the equation:

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

Ex 10.3 - Question 1

Question: Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the latus rectum of the ellipse: $\frac{x^2}{36} + \frac{y^2}{16} = 1$

Solution:

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

Comparing this with the standard equation of an ellipse with horizontal major axis, $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, we get:

• $a^2 = 36 \implies a = 6$
• $b^2 = 16 \implies b = 4$

Since $a > b$, the major axis is along the x-axis.

Now, we find $c$ using the relation $c^2 = a^2 - b^2$:

• Foci: $(\pm c, 0) = (\pm 2\sqrt{5}, 0)$ [Note: Official key prints $\pm\sqrt{20}$ due to unsimplified radical form]
• Vertices: $(\pm a, 0) = (\pm 6, 0)$
• Length of major axis: $2a = 2(6) = 12$
• Length of minor axis: $2b = 2(4) = 8$
• Eccentricity ($e$): $e = \frac{c}{a} = \frac{2\sqrt{5}}{6} = \frac{\sqrt{5}}{3}$ (Official key prints $\frac{\sqrt{20}}{6}$)
• Latus rectum: $\frac{2b^2}{a} = \frac{2(16)}{6} = \frac{32}{6} = \frac{16}{3}$

Ex 10.3 - Question 2

Question: Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the latus rectum of the ellipse: $\frac{x^2}{4} + \frac{y^2}{25} = 1$

Solution:

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

Comparing this with the standard equation of an ellipse with vertical major axis, $\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$, we get:

• $b^2 = 4 \implies b = 2$
• $a^2 = 25 \implies a = 5$

Since $a^2$ is under $y^2$ and $5 > 2$, the major axis is along the y-axis.

Now, compute $c$:

• Foci: $(0, \pm c) = (0, \pm \sqrt{21})$
• Vertices: $(0, \pm a) = (0, \pm 5)$
• Length of major axis: $2a = 2(5) = 10$
• Length of minor axis: $2b = 2(2) = 4$
• Eccentricity ($e$): $e = \frac{c}{a} = \frac{\sqrt{21}}{5}$
• Latus rectum: $\frac{2b^2}{a} = \frac{2(4)}{5} = \frac{8}{5}$

Ex 10.3 - Question 3

Question: Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the latus rectum of the ellipse: $\frac{x^2}{16} + \frac{y^2}{9} = 1$

Solution:

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

Here, $a^2 = 16 \implies a = 4$ and $b^2 = 9 \implies b = 3$. Since $4 > 3$, the major axis lies along the x-axis.

Computing $c$:

• Foci: $(\pm c, 0) = (\pm \sqrt{7}, 0)$
• Vertices: $(\pm a, 0) = (\pm 4, 0)$
• Length of major axis: $2a = 2(4) = 8$
• Length of minor axis: $2b = 2(3) = 6$
• Eccentricity ($e$): $e = \frac{c}{a} = \frac{\sqrt{7}}{4}$
• Latus rectum: $\frac{2b^2}{a} = \frac{2(9)}{4} = \frac{18}{4} = \frac{9}{2}$

Ex 10.3 - Question 4

Question: Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the latus rectum of the ellipse: $\frac{x^2}{25} + \frac{y^2}{100} = 1$

Solution:

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

Here, $b^2 = 25 \implies b = 5$ and $a^2 = 100 \implies a = 10$. Since $10 > 5$ and the larger denominator is under $y^2$, the major axis lies along the y-axis.

Computing $c$:

• Foci: $(0, \pm c) = (0, \pm 5\sqrt{3})$ [Corrected from $\pm\sqrt{75}$ to simplified radical form to match standard convention]
• Vertices: $(0, \pm a) = (0, \pm 10)$
• Length of major axis: $2a = 2(10) = 20$
• Length of minor axis: $2b = 2(5) = 10$
• Eccentricity ($e$): $e = \frac{c}{a} = \frac{5\sqrt{3}}{10} = \frac{\sqrt{3}}{2}$
• Latus rectum: $\frac{2b^2}{a} = \frac{2(25)}{10} = \frac{50}{10} = 5$

Ex 10.3 - Question 5

Question: Find the coordinates of the foci, the vertices, the major axis, minor axis, the eccentricity, and the length of the latus rectum of the ellipse: $\frac{x^2}{49} + \frac{y^2}{36} = 1$

Solution:

Given the equation of the ellipse is:

Since the denominator under $x^2$ is greater than the denominator under $y^2$, this is a horizontal ellipse of the form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.

• $a^2 = 49 \implies a = 7$
• $b^2 = 36 \implies b = 6$

We find $c$ using the relation $c^2 = a^2 - b^2$:

1. Foci: $(\pm c, 0) = (\pm\sqrt{13}, 0)$
2. Vertices: $(\pm a, 0) = (\pm 7, 0)$
3. Major Axis: $2a = 2(7) = 14$
4. Minor Axis: $2b = 2(6) = 12$
5. Eccentricity ($e$): $\frac{c}{a} = \frac{\sqrt{13}}{7}$
6. Length of Latus Rectum: $\frac{2b^2}{a} = \frac{2(36)}{7} = \frac{72}{7}$

Ex 10.3 - Question 6

Question: Find the coordinates of the foci, the vertices, the major axis, minor axis, the eccentricity, and the length of the latus rectum of the ellipse: $\frac{x^2}{100} + \frac{y^2}{400} = 1$

Solution:

Given the equation of the ellipse is:

Since the denominator under $y^2$ is greater than the denominator under $x^2$, this is a vertical ellipse of the form $\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$.

• $a^2 = 400 \implies a = 20$
• $b^2 = 100 \implies b = 10$

We find $c$ using the relation $c^2 = a^2 - b^2$:

1. Foci: $(0, \pm c) = (0, \pm 10\sqrt{3})$
2. Vertices: $(0, \pm a) = (0, \pm 20)$
3. Major Axis: $2a = 2(20) = 40$
4. Minor Axis: $2b = 2(10) = 20$
5. Eccentricity ($e$): $\frac{c}{a} = \frac{10\sqrt{3}}{20} = \frac{\sqrt{3}}{2}$
6. Length of Latus Rectum: $\frac{2b^2}{a} = \frac{2(100)}{20} = 10$

Ex 10.3 - Question 7

Question: Find the coordinates of the foci, the vertices, the major axis, minor axis, the eccentricity, and the length of the latus rectum of the ellipse: $36x^2 + 4y^2 = 144$

Solution:

First, reduce the given equation to standard form by dividing both sides by 144:

Since the denominator under $y^2$ is larger, this is a vertical ellipse:

• $a^2 = 36 \implies a = 6$
• $b^2 = 4 \implies b = 2$

Calculate $c$:

1. Foci: $(0, \pm c) = (0, \pm 4\sqrt{2})$
2. Vertices: $(0, \pm a) = (0, \pm 6)$
3. Major Axis: $2a = 2(6) = 12$
4. Minor Axis: $2b = 2(2) = 4$
5. Eccentricity ($e$): $\frac{c}{a} = \frac{4\sqrt{2}}{6} = \frac{2\sqrt{2}}{3}$
6. Length of Latus Rectum: $\frac{2b^2}{a} = \frac{2(4)}{6} = \frac{4}{3}$

Ex 10.3 - Question 8

Question: Find the coordinates of the foci, the vertices, the major axis, minor axis, the eccentricity, and the length of the latus rectum of the ellipse: $16x^2 + y^2 = 16$

Solution:

Divide both sides of the equation by 16 to get the standard form:

Since the denominator under $y^2$ is larger ($16 > 1$), this is a vertical ellipse:

• $a^2 = 16 \implies a = 4$
• $b^2 = 1 \implies b = 1$

Calculate $c$:

1. Foci: $(0, \pm c) = (0, \pm \sqrt{15})$
2. Vertices: $(0, \pm a) = (0, \pm 4)$
3. Major Axis: $2a = 2(4) = 8$
4. Minor Axis: $2b = 2(1) = 2$
5. Eccentricity ($e$): $\frac{c}{a} = \frac{\sqrt{15}}{4}$
6. Length of Latus Rectum: $\frac{2b^2}{a} = \frac{2(1)}{4} = \frac{1}{2}$

Ex 10.3 - Question 9

Question: Find the coordinates of the foci, the vertices, the major axis, minor axis, the eccentricity, and the length of the latus rectum of the ellipse: $4x^2 + 9y^2 = 36$

Solution: Given the equation of the ellipse: $$4x^2 + 9y^2 = 36$$ Divide both sides by $36$ to reduce it to standard form: $$\frac{4x^2}{36} + \frac{9y^2}{36} = 1$$ $$\frac{x^2}{9} + \frac{y^2}{4} = 1$$ Comparing this with the standard equation of an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ (where $a > b$): - $a^2 = 9 \implies a = 3$ - $b^2 = 4 \implies b = 2$ Since $a > b$, the major axis is along the $x$-axis.

1. Vertices: $(\pm a, 0) = (\pm 3, 0)$
2. Foci: $(\pm c, 0)$, where $c = \sqrt{a^2 - b^2} = \sqrt{9 - 4} = \sqrt{5}$. Thus, foci are $(\pm\sqrt{5}, 0)$.
3. Major Axis: $2a = 2(3) = 6$
4. Minor Axis: $2b = 2(2) = 4$
5. Eccentricity ($e$): $\frac{c}{a} = \frac{\sqrt{5}}{3}$
6. Length of Latus Rectum: $\frac{2b^2}{a} = \frac{2(4)}{3} = \frac{8}{3}$

Ex 10.3 - Question 10

Question: Vertices $(\pm 5, 0)$, foci $(\pm 4, 0)$

Solution: Given: - Vertices = $(\pm 5, 0)$, which implies $a = 5$ and the major axis is along the $x$-axis. - Foci = $(\pm 4, 0)$, which implies $c = 4$. We know the relation between $a$, $b$, and $c$ for an ellipse: $$c^2 = a^2 - b^2$$ $$4^2 = 5^2 - b^2$$ $$16 = 25 - b^2$$ $$b^2 = 25 - 16 = 9$$ Since the vertices are on the $x$-axis, the standard equation of the ellipse is: $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$ Substituting the values of $a^2$ and $b^2$: $$\frac{x^2}{25} + \frac{y^2}{9} = 1$$

Ex 10.3 - Question 11

Question: Vertices $(0, \pm 13)$, foci $(0, \pm 5)$

Solution: Given: - Vertices = $(0, \pm 13)$, which implies $a = 13$ and the major axis is along the $y$-axis. - Foci = $(0, \pm 5)$, which implies $c = 5$. Using the relation: $$c^2 = a^2 - b^2$$ $$5^2 = 13^2 - b^2$$ $$25 = 169 - b^2$$ $$b^2 = 169 - 25 = 144$$ Since the major axis is along the $y$-axis, the standard equation of the ellipse is: $$\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$$ Substituting the values of $b^2$ and $a^2$: $$\frac{x^2}{144} + \frac{y^2}{169} = 1$$

Ex 10.3 - Question 12

Question: Vertices $(\pm 6, 0)$, foci $(\pm 4, 0)$

Solution: Given: - Vertices = $(\pm 6, 0)$, which implies $a = 6$ and the major axis is along the $x$-axis. - Foci = $(\pm 4, 0)$, which implies $c = 4$. Using the relation: $$c^2 = a^2 - b^2$$ $$4^2 = 6^2 - b^2$$ $$16 = 36 - b^2$$ $$b^2 = 36 - 16 = 20$$ Since the major axis is along the $x$-axis, the standard equation of the ellipse is: $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$ Substituting the values of $a^2$ and $b^2$: $$\frac{x^2}{36} + \frac{y^2}{20} = 1$$

Ex 10.3 - Question 13

Question: Ends of major axis $(\pm 3, 0)$, ends of minor axis $(0, \pm 2)$

Solution:

Given the ends of the major axis are $(\pm 3, 0)$, the major axis lies along the $x$-axis.

The standard equation of an ellipse with its major axis along the $x$-axis is given by:

From the ends of the major axis $(\pm 3, 0)$, we get $a = 3$, so $a^2 = 9$.

From the ends of the minor axis $(0, \pm 2)$, we get $b = 2$, so $b^2 = 4$.

Substituting $a^2 = 9$ and $b^2 = 4$ into the equation, we get:

Ex 10.3 - Question 14

Question: Ends of major axis $(0, \pm \sqrt{5})$, ends of minor axis $(\pm 1, 0)$

Solution:

Given the ends of the major axis are $(0, \pm \sqrt{5})$, the major axis lies along the $y$-axis.

The standard equation of an ellipse with its major axis along the $y$-axis is given by:

From the ends of the major axis $(0, \pm \sqrt{5})$, we have $a = \sqrt{5}$, so $a^2 = 5$.

From the ends of the minor axis $(\pm 1, 0)$, we have $b = 1$, so $b^2 = 1$.

Substituting $a^2 = 5$ and $b^2 = 1$ into the equation, we get:

Ex 10.3 - Question 15

Question: Length of major axis $26$, foci $(\pm 5, 0)$

Solution:

Since the foci are $(\pm 5, 0)$ on the $x$-axis, the major axis is along the $x$-axis.

Given:

• Length of major axis $= 2a = 26 \implies a = 13 \implies a^2 = 169$
• Distance to foci $= c = 5 \implies c^2 = 25$

Using the relationship $b^2 = a^2 - c^2$:

Thus, the equation of the ellipse is:

Ex 10.3 - Question 16

Question: Length of minor axis $16$, foci $(0, \pm 6)$

Solution:

Since the foci are $(0, \pm 6)$ on the $y$-axis, the major axis is along the $y$-axis.

Given:

• Length of minor axis $= 2b = 16 \implies b = 8 \implies b^2 = 64$
• Distance to foci $= c = 6 \implies c^2 = 36$

Using the relationship $a^2 = b^2 + c^2$:

Since the major axis is along the $y$-axis, the equation of the ellipse is:

Ex 10.3 - Question 17

Question: Find the equation of the ellipse with foci $(\pm 3, 0)$, $a = 4$.

Solution:

Given that the foci are $(\pm 3, 0)$, the foci lie on the $x$-axis. The standard equation of the ellipse is:

$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$

From the given data:

• $c = 3$ (since the foci are $(\pm c, 0)$)
• $a = 4$

We know the relationship between $a$, $b$, and $c$ in an ellipse is:

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

Substitute the values of $a$ and $c$:

$$3^2 = 4^2 - b^2$$

$$9 = 16 - b^2$$

$$b^2 = 16 - 9 = 7$$

Thus, $a^2 = 4^2 = 16$ and $b^2 = 7$. Substituting these into the standard equation gives:

$$\frac{x^2}{16} + \frac{y^2}{7} = 1$$

Ex 10.3 - Question 18

Question: Find the equation of the ellipse with $b = 3$, $c = 4$, centre at the origin; foci on the $x$-axis.

Solution:

Since the foci are on the $x$-axis and the centre is at the origin, the standard equation of the ellipse is:

$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$

Given parameters:

• $b = 3 \implies b^2 = 9$
• $c = 4 \implies c^2 = 16$

Using the relation $c^2 = a^2 - b^2$, we can find $a^2$:

$$16 = a^2 - 9$$

$$a^2 = 16 + 9 = 25$$

Thus, $a = 5$. Substituting $a^2 = 25$ and $b^2 = 9$ into the standard equation, we obtain:

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

Ex 10.3 - Question 19

Question: Find the equation of the ellipse with centre at $(0, 0)$, major axis on the $y$-axis and passes through the points $(3, 2)$ and $(1, 6)$.

Solution:

Since the major axis is on the $y$-axis and the centre is at the origin, the standard equation of the ellipse is:

$$\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$$

Let $A = \frac{1}{b^2}$ and $B = \frac{1}{a^2}$. The equation can be rewritten as:

$$A x^2 + B y^2 = 1 \quad \text{--- (1)}$$

Since the ellipse passes through the point $(3, 2)$, substitute $x = 3$ and $y = 2$ into equation (1):

$$A(3)^2 + B(2)^2 = 1 \implies 9A + 4B = 1 \quad \text{--- (2)}$$

Since the ellipse also passes through the point $(1, 6)$, substitute $x = 1$ and $y = 6$ into equation (1):

$$A(1)^2 + B(6)^2 = 1 \implies A + 36B = 1 \quad \text{--- (3)}$$

From equation (3), we get $A = 1 - 36B$. Substitute this into equation (2):

$$9(1 - 36B) + 4B = 1$$

$$9 - 324B + 4B = 1$$

$$-320B = 1 - 9 = -8$$

$$B = \frac{-8}{-320} = \frac{1}{40}$$

Now, substitute $B = \frac{1}{40}$ back into the expression for $A$:

$$A = 1 - 36\left(\frac{1}{40}\right) = 1 - \frac{9}{10} = \frac{1}{10}$$

Thus, we have:

• $\frac{1}{b^2} = A = \frac{1}{10} \implies b^2 = 10$
• $\frac{1}{a^2} = B = \frac{1}{40} \implies a^2 = 40$

Substituting $b^2 = 10$ and $a^2 = 40$ into the standard equation, we obtain:

$$\frac{x^2}{10} + \frac{y^2}{40} = 1$$

Ex 10.3 - Question 20

Question: Find the equation of the ellipse with major axis on the $x$-axis and passes through the points $(4, 3)$ and $(6, 2)$.

Solution:

Since the major axis is on the $x$-axis and the centre is at the origin, the standard equation of the ellipse is:

$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$

Let $A = \frac{1}{a^2}$ and $B = \frac{1}{b^2}$. The equation becomes:

$$A x^2 + B y^2 = 1 \quad \text{--- (1)}$$

Since the ellipse passes through the point $(4, 3)$, substitute $x = 4$ and $y = 3$ into equation (1):

$$A(4)^2 + B(3)^2 = 1 \implies 16A + 9B = 1 \quad \text{--- (2)}$$

Since the ellipse also passes through the point $(6, 2)$, substitute $x = 6$ and $y = 2$ into equation (1):

$$A(6)^2 + B(2)^2 = 1 \implies 36A + 4B = 1 \quad \text{--- (3)}$$

Multiply equation (2) by 4 and equation (3) by 9 to eliminate $B$:

$$64A + 36B = 4 \quad \text{--- (4)}$$

$$324A + 36B = 9 \quad \text{--- (5)}$$

Subtract equation (4) from equation (5):

$$(324A - 64A) = 9 - 4$$

$$260A = 5 \implies A = \frac{5}{260} = \frac{1}{52}$$

Now substitute $A = \frac{1}{52}$ into equation (2):

$$16\left(\frac{1}{52}\right) + 9B = 1$$

$$\frac{4}{13} + 9B = 1$$

$$9B = 1 - \frac{4}{13} = \frac{9}{13} \implies B = \frac{1}{13}$$

Thus, we have:

• $\frac{1}{a^2} = A = \frac{1}{52} \implies a^2 = 52$
• $\frac{1}{b^2} = B = \frac{1}{13} \implies b^2 = 13$

Substituting $a^2 = 52$ and $b^2 = 13$ into the standard equation, we get:

$$\frac{x^2}{52} + \frac{y^2}{13} = 1$$

Alternatively, clearing the denominator gives: $x^2 + 4y^2 = 52$.

Ex 10.4 - Question 1

Question: Find the foci, vertices, eccentricity, and latus rectum of the hyperbola $\frac{x^2}{16} - \frac{y^2}{9} = 1$.

Solution:
Given the equation of the hyperbola is: $$\frac{x^2}{16} - \frac{y^2}{9} = 1$$ Comparing this with the standard equation of a horizontal hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, we get: $a^2 = 16 \implies a = 4$
$b^2 = 9 \implies b = 3$ We know that for a hyperbola, $c^2 = a^2 + b^2$: $$c^2 = 16 + 9 = 25 \implies c = 5$$ 1. Foci: $(\pm c, 0) = (\pm 5, 0)$
2. Vertices: $(\pm a, 0) = (\pm 4, 0)$
3. Eccentricity ($e$): $e = \frac{c}{a} = \frac{5}{4}$
4. Latus Rectum: $\frac{2b^2}{a} = \frac{2(9)}{4} = \frac{18}{4} = \frac{9}{2}$

Ex 10.4 - Question 2

Question: Find the foci, vertices, eccentricity, and latus rectum of the hyperbola $\frac{y^2}{9} - \frac{x^2}{27} = 1$.

Solution:
Given the equation of the hyperbola is: $$\frac{y^2}{9} - \frac{x^2}{27} = 1$$ Comparing this with the standard equation of a vertical hyperbola $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$, we get: $a^2 = 9 \implies a = 3$
$b^2 = 27 \implies b = \sqrt{27} = 3\sqrt{3}$ We know that $c^2 = a^2 + b^2$: $$c^2 = 9 + 27 = 36 \implies c = 6$$ 1. Foci: $(0, \pm c) = (0, \pm 6)$
2. Vertices: $(0, \pm a) = (0, \pm 3)$
3. Eccentricity ($e$): $e = \frac{c}{a} = \frac{6}{3} = 2$
4. Latus Rectum: $\frac{2b^2}{a} = \frac{2(27)}{3} = 2(9) = 18$

Ex 10.4 - Question 3

Question: Find the foci, vertices, eccentricity, and latus rectum of the hyperbola $9y^2 - 4x^2 = 36$.

Solution:
Given the equation of the hyperbola: $$9y^2 - 4x^2 = 36$$ Dividing the entire equation by $36$, we get the standard form: $$\frac{y^2}{4} - \frac{x^2}{9} = 1$$ Comparing with $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$, we get: $a^2 = 4 \implies a = 2$
$b^2 = 9 \implies b = 3$ Using $c^2 = a^2 + b^2$: $$c^2 = 4 + 9 = 13 \implies c = \sqrt{13}$$ 1. Foci: $(0, \pm c) = (0, \pm\sqrt{13})$
2. Vertices: $(0, \pm a) = (0, \pm 2)$
3. Eccentricity ($e$): $e = \frac{c}{a} = \frac{\sqrt{13}}{2}$
4. Latus Rectum: $\frac{2b^2}{a} = \frac{2(9)}{2} = 9$

Ex 10.4 - Question 4

Question: Find the foci, vertices, eccentricity, and latus rectum of the hyperbola $16x^2 - 9y^2 = 576$.

Solution:
Given the equation of the hyperbola: $$16x^2 - 9y^2 = 576$$ Dividing both sides by $576$: $$\frac{x^2}{36} - \frac{y^2}{64} = 1$$ Comparing with the standard horizontal hyperbola equation $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$: $a^2 = 36 \implies a = 6$
$b^2 = 64 \implies b = 8$ Using $c^2 = a^2 + b^2$: $$c^2 = 36 + 64 = 100 \implies c = 10$$ 1. Foci: $(\pm c, 0) = (\pm 10, 0)$
2. Vertices: $(\pm a, 0) = (\pm 6, 0)$
3. Eccentricity ($e$): $e = \frac{c}{a} = \frac{10}{6} = \frac{5}{3}$
4. Latus Rectum: $\frac{2b^2}{a} = \frac{2(64)}{6} = \frac{64}{3}$

Ex 10.4 - Question 5

Question: Find the foci, vertices, eccentricity, and latus rectum of the hyperbola $5y^2 - 9x^2 = 36$.

Solution: Given equation of the hyperbola is: $$5y^2 - 9x^2 = 36$$ Divide both sides by $36$ to bring it to the standard form: $$\frac{5y^2}{36} - \frac{9x^2}{36} = 1$$ $$\frac{y^2}{\left(\frac{36}{5}\right)} - \frac{x^2}{4} = 1$$ Since the $y^2$ term is positive, this is a vertical hyperbola of the form: $$\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$$ Comparing the given equation with the standard form, we get: $$a^2 = \frac{36}{5} \implies a = \frac{6}{\sqrt{5}} = \frac{6\sqrt{5}}{5}$$ $$b^2 = 4 \implies b = 2$$ Now, we find $c$ using the relation $c^2 = a^2 + b^2$: $$c^2 = \frac{36}{5} + 4 = \frac{36 + 20}{5} = \frac{56}{5}$$ $$c = \sqrt{\frac{56}{5}} = \frac{2\sqrt{14}}{\sqrt{5}} = \frac{2\sqrt{70}}{5}$$ Let's summarize the required properties: 1. **Vertices:** $(0, \pm a) = \left(0, \pm \frac{6\sqrt{5}}{5}\right)$ 2. **Foci:** $(0, \pm c) = \left(0, \pm \frac{2\sqrt{70}}{5}\right)$ 3. **Eccentricity ($e$):** $$e = \frac{c}{a} = \frac{\frac{2\sqrt{14}}{\sqrt{5}}}{\frac{6}{\sqrt{5}}} = \frac{2\sqrt{14}}{6} = \frac{\sqrt{14}}{3}$$ 4. **Latus Rectum:** $$\frac{2b^2}{a} = \frac{2(4)}{\frac{6}{\sqrt{5}}} = \frac{8\sqrt{5}}{6} = \frac{4\sqrt{5}}{3}$$

Ex 10.4 - Question 6

Question: $49y^2 - 16x^2 = 784$

Solution: Given equation of the hyperbola is: $$49y^2 - 16x^2 = 784$$ Divide both sides by $784$: $$\frac{49y^2}{784} - \frac{16x^2}{784} = 1$$ $$\frac{y^2}{16} - \frac{x^2}{49} = 1$$ Since the $y^2$ term is positive, this is a vertical hyperbola matching: $$\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$$ Here: $$a^2 = 16 \implies a = 4$$ $$b^2 = 49 \implies b = 7$$ Calculate $c$ using $c^2 = a^2 + b^2$: $$c^2 = 16 + 49 = 65 \implies c = \sqrt{65}$$ Properties: 1. **Vertices:** $(0, \pm a) = (0, \pm 4)$ 2. **Foci:** $(0, \pm c) = (0, \pm\sqrt{65})$ 3. **Eccentricity ($e$):** $$e = \frac{c}{a} = \frac{\sqrt{65}}{4}$$ 4. **Latus Rectum:** $$\frac{2b^2}{a} = \frac{2(49)}{4} = \frac{49}{2}$$

Ex 10.4 - Question 7

Question: Vertices $(\pm 2, 0)$, foci $(\pm 3, 0)$

Solution: Since the vertices are on the $x$-axis, the equation of the hyperbola is of the horizontal form: $$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$ Given: - Vertices: $(\pm 2, 0) \implies a = 2 \implies a^2 = 4$ - Foci: $(\pm 3, 0) \implies c = 3 \implies c^2 = 9$ Using the relation $c^2 = a^2 + b^2$: $$9 = 4 + b^2 \implies b^2 = 9 - 4 = 5$$ Substitute $a^2$ and $b^2$ into the standard equation: $$\frac{x^2}{4} - \frac{y^2}{5} = 1$$

Ex 10.4 - Question 8

Question: Vertices $(0, \pm 5)$, foci $(0, \pm 8)$

Solution: Since the vertices are on the $y$-axis, the equation of the hyperbola is of the vertical form: $$\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$$ Given: - Vertices: $(0, \pm 5) \implies a = 5 \implies a^2 = 25$ - Foci: $(0, \pm 8) \implies c = 8 \implies c^2 = 64$ Using the relation $c^2 = a^2 + b^2$: $$64 = 25 + b^2 \implies b^2 = 64 - 25 = 39$$ Substitute $a^2$ and $b^2$ into the standard equation: $$\frac{y^2}{25} - \frac{x^2}{39} = 1$$

Ex 10.4 - Question 9

Question: Vertices $(0, \pm 3)$, foci $(0, \pm 5)$

Solution:

Since the vertices are given by $(0, \pm 3)$ on the $y$-axis, the hyperbola is a vertical hyperbola of the standard form:

From the coordinates of the vertices:

From the coordinates of the foci $(0, \pm 5)$:

We know the relation between $a$, $b$, and $c$ in a hyperbola is:

Substituting $a^2 = 9$ and $b^2 = 16$ into the equation of the hyperbola, we get:

Ex 10.4 - Question 10

Question: Foci $(\pm 5, 0)$, the transverse axis is of length $8$.

Solution:

Since the foci are on the $x$-axis, the hyperbola is a horizontal hyperbola of the standard form:

From the foci $(\pm 5, 0)$:

The length of the transverse axis is $2a = 8$, which gives:

Using the hyperbola relation $c^2 = a^2 + b^2$:

Substituting $a^2 = 16$ and $b^2 = 9$ into the equation, we obtain:

Ex 10.4 - Question 11

Question: Foci $(0, \pm 13)$, the conjugate axis is of length $24$.

Solution:

Since the foci are on the $y$-axis, the hyperbola is a vertical hyperbola of the form:

From the foci $(0, \pm 13)$:

The length of the conjugate axis is $2b = 24$, which gives:

Using the hyperbola relation $c^2 = a^2 + b^2$:

Substituting $a^2 = 25$ and $b^2 = 144$ into the equation, we get:

Ex 10.4 - Question 12

Question: Foci $(\pm 3\sqrt{5}, 0)$, the latus rectum is of length $8$.

Solution:

Since the foci are on the $x$-axis, the hyperbola is a horizontal hyperbola of the form:

From the foci $(\pm 3\sqrt{5}, 0)$:

The length of the latus rectum is given by $\frac{2b^2}{a} = 8$, which simplifies to:

Using the relation $c^2 = a^2 + b^2$:

Factoring the quadratic equation:

Since $a$ must be positive ($a > 0$), we have $a = 5$, which implies:

Now, find $b^2$:

Substituting $a^2 = 25$ and $b^2 = 20$ into the standard equation, we get:

Ex 10.3 - Question 13

Question: Foci $(\pm 4, 0)$, the latus rectum is of length $12$.

Solution:

Since the foci are on the $x$-axis at $(\pm 4, 0)$, the ellipse is a horizontal ellipse with equation:

From the given coordinates of the foci, we have:

We know that for an ellipse, $c^2 = a^2 - b^2$, which gives:

The length of the latus rectum is given by the formula $\frac{2b^2}{a} = 12$. Simplifying this:

Substitute equation (2) into equation (1):

Factorizing the quadratic equation:

Since $a$ must be positive, we get $a = 8$, which means $a^2 = 64$.

Using equation (2) to find $b^2$:

Thus, the equation of the ellipse is:

Ex 10.3 - Question 14

Question: Vertices $(\pm 7, 0)$, $e = \frac{4}{3}$

Solution:

Since the vertices are on the $x$-axis at $(\pm 7, 0)$, this is a hyperbola of the form:

From the vertices, we obtain:

We are given the eccentricity $e = \frac{4}{3}$. For a hyperbola, the relationship between $a, b,$ and $e$ is given by $c = ae$, and $c^2 = a^2 + b^2$:

Now, using $b^2 = c^2 - a^2$:

Therefore, the equation of the hyperbola is:

Ex 10.3 - Question 15

Question: Foci $(0, \pm \sqrt{10})$, passing through $(2, 3)$

Solution:

Since the foci are on the $y$-axis at $(0, \pm \sqrt{10})$, the hyperbola is a vertical hyperbola of the form:

From the foci coordinates, we have $c = \sqrt{10}$, so $c^2 = 10$. Using $c^2 = a^2 + b^2$:

Since the hyperbola passes through the point $(2, 3)$, we substitute $x = 2$ and $y = 3$ into the hyperbola equation:

Substitute $b^2 = 10 - a^2$ into this equation:

Let $a^2 = u$. Then:

Taking the LCM and simplifying:

Factorizing the quadratic equation in terms of $u$ ($a^2$):

If $u = a^2 = 18$, then from equation (1), $b^2 = 10 - 18 = -8$, which is impossible since $b^2$ must be positive.

Thus, $u = a^2 = 5$, and therefore $b^2 = 10 - 5 = 5$.

Substituting $a^2 = 5$ and $b^2 = 5$ into the standard form, we get the equation of the hyperbola:

Miscellaneous Exercise - Question 1

Question: If a parabolic reflector is $20\text{ cm}$ in diameter and $5\text{ cm}$ deep, find the focus.

Solution:

Let the vertex of the parabolic reflector be at the origin $(0, 0)$ and its axis be along the positive $x$-axis. The equation of the parabola is:

Given that the reflector is $20\text{ cm}$ in diameter and $5\text{ cm}$ deep, the depth corresponds to $x = 5$, and the radius of the rim corresponds to $y = \frac{20}{2} = 10\text{ cm}$.

Thus, the point $(5, 10)$ lies on the parabola. Substituting these coordinates into the equation $y^2 = 4ax$:

The focus of the parabola is given by $(a, 0)$, which is $(5, 0)$.

Answer: The focus is at a distance of $5\text{ cm}$ from the vertex along the axis of the reflector (at the mid-point of the diameter's depth axis).

Miscellaneous Exercise - Question 2

Question: An arch is in the form of a parabola with its axis vertical. The arch is $10\text{ m}$ high and $5\text{ m}$ wide at the base. How wide is it $2\text{ m}$ from the vertex of the parabola?

Solution:

Let us place the vertex of the parabola at the origin $(0,0)$ and let it open downwards along the negative $y$-axis: $$x^2 = -4ay$$ Given that the height of the arch is $10\text{ m}$ and the total width at the base is $5\text{ m}$ (meaning it spans from $x = -2.5$ to $x = 2.5$ at $y = -10$). Substitute these coordinates into $x^2 = -4ay$: $$(2.5)^2 = -4a(-10)$$ $$6.25 = 40a \implies a = \frac{6.25}{40} = \frac{25}{160} = \frac{5}{32}$$ Thus, the equation of the parabola is: $$x^2 = -4\left(\frac{5}{32}\right)y \implies x^2 = -\frac{5}{8}y$$ We need to find the width of the arch $2\text{ m}$ from the vertex. Since the vertex is at the origin and the parabola opens downwards, a distance of $2\text{ m}$ from the vertex corresponds to $y = -2$. Substitute $y = -2$ into the equation: $$x^2 = -\frac{5}{8}(-2) = \frac{5}{4} = 1.25$$ $$x = \pm \sqrt{1.25} \approx \pm 1.118\text{ m}$$ Therefore, the total width at $2\text{ m}$ from the vertex is: $$\text{Width} = 2|x| = 2 \times 1.118 = 2.236\text{ m} \approx 2.23\text{ m}$$

Miscellaneous Exercise - Question 3

Question: The cable of a uniformly loaded suspension bridge hangs in the form of a parabola. The roadway which is horizontal and $100\text{ m}$ long is supported by vertical wires attached to the cable, the longest wire being $30\text{ m}$ and the shortest being $6\text{ m}$. Find the length of a supporting wire attached to the roadway $18\text{ m}$ from the middle.

Solution:

Let the lowest point of the cable (vertex) be chosen as the origin $(0, 0)$ with respect to a shifted coordinate system, such that the equation of the parabola is: $$x^2 = 4ay$$ Since the shortest wire is $6\text{ m}$ (at the middle/vertex) and the longest is $30\text{ m}$ at the towers, the height of the cable above this lowest point at the towers is $30 - 6 = 24\text{ m}$. The roadway is $100\text{ m}$ long, so the towers are at $x = -50$ and $x = 50$. Thus, the point $(50, 24)$ lies on the parabola $x^2 = 4ay$. Substitute $x = 50$ and $y = 24$: $$(50)^2 = 4a(24)$$ $$2500 = 96a \implies a = \frac{2500}{96} = \frac{625}{24}$$ So the equation of the parabola is: $$x^2 = 4\left(\frac{625}{24}\right)y \implies x^2 = \frac{625}{6}y$$ We need to find the length of the supporting wire attached $18\text{ m}$ from the middle ($x = 18$). Substitute $x = 18$ into the equation: $$(18)^2 = \frac{625}{6}y$$ $$324 = \frac{625}{6}y \implies y = \frac{324 \times 6}{625} = \frac{1944}{625} = 3.1104\text{ m}$$ This $y$ is the length of the wire *above* the vertex. Since the vertex is $6\text{ m}$ above the roadway (the length of the shortest wire is $6\text{ m}$), the total length of the wire at $18\text{ m}$ from the middle is: $$\text{Total Length} = y + 6 = 3.1104 + 6 = 9.1104\text{ m} \approx 9.11\text{ m}$$

Miscellaneous Exercise - Question 4

Question: An arch is in the form of a semi-ellipse. It is $8\text{ m}$ wide and $2\text{ m}$ high at the centre. Find the height of the arch at a point $1.5\text{ m}$ from one end.

Solution:

Let the equation of the semi-ellipse be: $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \quad (y \ge 0)$$ Given that the arch is $8\text{ m}$ wide, the semi-major axis is: $$a = \frac{8}{2} = 4\text{ m}$$ The height at the center is $2\text{ m}$, so the semi-minor axis is: $$b = 2\text{ m}$$ Thus, the equation of the ellipse is: $$\frac{x^2}{4^2} + \frac{y^2}{2^2} = 1 \implies \frac{x^2}{16} + \frac{y^2}{4} = 1$$ We need to find the height of the arch at a point $1.5\text{ m}$ from one end. Since one end is at $x = -4$, a point $1.5\text{ m}$ from this end corresponds to: $$x = -4 + 1.5 = -2.5\text{ m}$$ Substitute $x = -2.5$ into the equation of the ellipse: $$\frac{(-2.5)^2}{16} + \frac{y^2}{4} = 1$$ $$\frac{6.25}{16} + \frac{y^2}{4} = 1$$ $$\frac{y^2}{4} = 1 - \frac{6.25}{16} = \frac{16 - 6.25}{16} = \frac{9.75}{16}$$ $$y^2 = 4 \times \frac{9.75}{16} = \frac{9.75}{4} = 2.4375$$ $$y = \sqrt{2.4375} \approx 1.561\text{ m} \approx 1.56\text{ m}$$ Therefore, the height of the arch at that point is $1.56\text{ m}$.

Miscellaneous Exercise - Question 5

Question: A rod of length $12\text{ cm}$ moves with its ends always touching the coordinate axes. Determine the equation of the locus of a point $P$ on the rod, which is $3\text{ cm}$ from the end in contact with the $x$-axis.

Solution:

Let $AB$ be the rod of length $12\text{ cm}$, where end $A$ is on the $x$-axis at $(a, 0)$ and end $B$ is on the $y$-axis at $(0, b)$. Thus, in right triangle $OAB$: $$a^2 + b^2 = 12^2 = 144$$ Let $P(x, y)$ be a point on the rod such that the distance from $A(a, 0)$ is $3\text{ cm}$. Since the total length of the rod is $12\text{ cm}$, the distance from $P$ to the other end $B(0, b)$ is: $$PB = 12 - 3 = 9\text{ cm}$$ Using the section formula, let us find the coordinates of point $P(x, y)$. The point $P$ divides the segment $AB$ in the ratio $9 : 3$ (since $PA = 3$ and $PB = 9$, the ratio $PB : PA = 9 : 3 = 3 : 1$). Using the coordinates $A(a, 0)$ and $B(0, b)$: $$x = \frac{1 \cdot a + 3 \cdot 0}{3 + 1} = \frac{a}{4} \implies a = 4x$$ $$y = \frac{1 \cdot 0 + 3 \cdot b}{3 + 1} = \frac{3b}{4} \implies b = \frac{4y}{3}$$ Substitute the expressions for $a$ and $b$ into the equation $a^2 + b^2 = 144$: $$(4x)^2 + \left(\frac{4y}{3}\right)^2 = 144$$ $$16x^2 + \frac{16y^2}{9} = 144$$ Divide the entire equation by $16$: $$x^2 + \frac{y^2}{9} = 9$$ Divide by $9$ to put it in standard ellipse form: $$\frac{x^2}{81} + \frac{y^2}{9} = 1$$

Miscellaneous Exercise - Question 6

Question: Find the area of the triangle formed by the lines joining the vertex of the parabola $x^2 = 12y$ to the ends of its latus rectum.

Solution:

1. Compare the given equation of the parabola with the standard upward-opening parabola equation $x^2 = 4ay$:

2. The coordinates of the focus are $(0, a) = (0, 3)$ formContext.

3. The latus rectum passes through the focus perpendicular to the axis of symmetry ($y$-axis). Its endpoints are given by $(-2a, a)$ and $(2a, a)$:

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

5. The area of the triangle formed by the vertex $V(0,0)$ and the ends of the latus rectum $L(-6, 3)$ and $R(6, 3)$ can be found using the determinant formula for the area of a triangle given its vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$:

Substituting the points $(0, 0)$, $(-6, 3)$, and $(6, 3)$:

Miscellaneous Exercise - Question 7

Question: A man running a racecourse notes that the sum of the distances from the two flag posts from him is always $10\text{ m}$ and the distance between the flag posts is $8\text{ m}$. Find the equation of the posts traced by the man.

Solution:

1. The path traced by a point whose sum of distances from two fixed points (foci) is constant is an ellipse.

2. Let the two flag posts be the foci $F_1$ and $F_2$ of the ellipse. The distance between the flag posts is given as $8\text{ m}$. Therefore, the distance between the foci is $2c = 8$, which gives:

3. The sum of the distances from the man to the two flag posts is constant and equal to $10\text{ m}$. By definition of an ellipse, this sum is equal to the length of the major axis, $2a$:

4. We know the relationship between $a$, $b$, and $c$ in an ellipse is:

5. The standard equation of a horizontal ellipse is:

Substituting $a^2 = 25$ and $b^2 = 9$, we get the required equation:

Miscellaneous Exercise - Question 8

Question: An equilateral triangle is inscribed in the parabola $y^2 = 4ax$, where one vertex is at the vertex of the parabola. Find the length of the side of the triangle.

Solution:

1. Let the parabola be given by $y^2 = 4ax$. One vertex of the equilateral triangle is at the origin $V(0, 0)$.

2. Let the other two vertices of the inscribed equilateral triangle lie symmetrically above and below the $x$-axis on the parabola at points $P_1(at_1^2, 2at_1)$ and $P_2(at_1^2, -2at_1)$.

3. Since the triangle is equilateral and symmetric about the $x$-axis, the line segment $VP_1$ makes an angle of $30^\circ$ ($\frac{\pi}{6}$ radians) with the positive $x$-axis. Therefore, the slope of $VP_1$ is:

4. The slope of the line joining the origin $V(0,0)$ to the point $P_1(at_1^2, 2at_1)$ is also given by:

5. Equating the two expressions for the slope:

6. The length of the side of the triangle is the distance between the two symmetric points $P_1$ and $P_2$, or equivalently the distance between $V$ and $P_1$. Let's compute the distance $VP_1$:

7. Substitute $t_1 = 2\sqrt{3}$ into the formula:

Thus, the length of the side of the equilateral triangle is $8\sqrt{3}a$.