Conic Sections Class 11 Mathematics NCERT Solutions

Q1Exercise 10.1

In each of the following Exercises 1 to 5, find the equation of the circle with centre $(0,2)$ and radius 2

Solution

Given: Centre of the circle

(h, k) = (0, 2)

Radius of the circle

r = 2

To Find: The equation of the circle.

Formula: The equation of a circle with centre

(h, k)

and radius

r

is given by:

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

Solution: Substituting the given values into the formula:

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

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

Expanding the equation:

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

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

Final Answer: The equation of the circle is

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

or

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

.

Q2Exercise 10.1

centre $(-2,3)$ and radius 4

Solution

Given: Centre of the circle

(h, k) = (-2, 3)

Radius of the circle

r = 4

To Find: The equation of the circle.

Formula: The equation of a circle with centre

(h, k)

and radius

r

is given by:

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

Solution: Substituting the given values into the formula:

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

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

Expanding the equation:

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

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

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

Final Answer: The equation of the circle is

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

or

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

.

Q3Exercise 10.1

centre $\left(\frac{1}{2}, \frac{1}{4}\right)$ and radius $\frac{1}{12}$

Solution

Given: Centre of the circle

(h, k) = \left(\frac{1}{2}, \frac{1}{4}\right)

Radius of the circle

r = \frac{1}{12}

To Find: The equation of the circle.

Formula: The equation of a circle with centre

(h, k)

and radius

r

is given by:

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

Solution: Substituting the given values into the formula:

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

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

Expanding the equation:

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

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

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

x^2 + y^2 - x - \frac{1}{2}y + \frac{5}{16} = \frac{1}{144}

Multiplying by 144 to clear fractions:

144x^2 + 144y^2 - 144x - 72y + 45 = 1

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

Dividing by 4:

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

Final Answer: The equation of the circle is

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

or

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

.

Q4Exercise 10.1

centre $(1,1)$ and radius $\sqrt{2}$

Solution

Given: Centre of the circle

(h, k) = (1, 1)

Radius of the circle

r = \sqrt{2}

To Find: The equation of the circle.

Formula: The equation of a circle with centre

(h, k)

and radius

r

is given by:

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

Solution: Substituting the given values into the formula:

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

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

Expanding the equation:

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

x^2 + y^2 - 2x - 2y + 2 = 2

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

Final Answer: The equation of the circle is

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

or

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

.

Q5Exercise 10.1

centre $(-a,-b)$ and radius $\sqrt{a^{2}-b^{2}}$

Solution

Given: Centre of the circle

(h, k) = (-a, -b)

Radius of the circle

r = \sqrt{a^2 - b^2}

To Find: The equation of the circle.

Formula: The equation of a circle with centre

(h, k)

and radius

r

is given by:

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

Solution: Substituting the given values into the formula:

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

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

Expanding the equation:

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

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

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

Final Answer: The equation of the circle is

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

or

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

.

Q6Exercise 10.1

In each of the following Exercises 6 to 9, find the centre and radius of the circles. 6. $(x+5)^{2}+(y-3)^{2}=36$

Solution

Given: The equation of the circle is

(x + 5)^2 + (y - 3)^2 = 36

.

To Find: The centre and radius of the circle.

Formula: The standard equation of a circle is

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

, where the centre is

(h, k)

and the radius is

r

.

Solution: We can rewrite the given equation to match the standard form:

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

Comparing this with the standard equation, we get:

h = -5

k = 3

r = 6

Final Answer: The centre of the circle is

(-5, 3)

and the radius is 6.

Q7Exercise 10.1

$x^{2}+y^{2}-4 x-8 y-45=0$

Solution

Given: The equation of the circle is

x^2 + y^2 - 4x - 8y - 45 = 0

.

To Find: The centre and radius of the circle.

Formula: The standard equation of a circle is

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

, where the centre is

(h, k)

and the radius is

r

.

Solution: We convert the given equation to the standard form by completing the square. Rearrange the terms:

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

Complete the square for the x-terms and y-terms:

(x^2 - 4x + (\frac{-4}{2})^2) + (y^2 - 8y + (\frac{-8}{2})^2) = 45 + (\frac{-4}{2})^2 + (\frac{-8}{2})^2

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

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

Rewrite in the standard form:

(x - 2)^2 + (y - 4)^2 = (\sqrt{65})^2

Comparing this with the standard equation, we get:

h = 2

k = 4

r = \sqrt{65}

Final Answer: The centre of the circle is

(2, 4)

and the radius is

\sqrt{65}

.

Q8Exercise 10.1

$x^{2}+y^{2}-8 x+10 y-12=0$

Solution

Given: The equation of the circle is

x^2 + y^2 - 8x + 10y - 12 = 0

.

To Find: The centre and radius of the circle.

Formula: The standard equation of a circle is

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

, where the centre is

(h, k)

and the radius is

r

.

Solution: We convert the given equation to the standard form by completing the square. Rearrange the terms:

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

Complete the square for the x-terms and y-terms:

(x^2 - 8x + (\frac{-8}{2})^2) + (y^2 + 10y + (\frac{10}{2})^2) = 12 + (\frac{-8}{2})^2 + (\frac{10}{2})^2

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

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

Rewrite in the standard form:

(x - 4)^2 + (y - (-5))^2 = (\sqrt{53})^2

Comparing this with the standard equation, we get:

h = 4

k = -5

r = \sqrt{53}

Final Answer: The centre of the circle is

(4, -5)

and the radius is

\sqrt{53}

.

Q9Exercise 10.1

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

Solution

Given: The equation of the circle is

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

.

To Find: The centre and radius of the circle.

Formula: The standard equation of a circle is

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

, where the centre is

(h, k)

and the radius is

r

.

Solution: First, divide the entire equation by 2 to make the coefficients of

x^2

and

y^2

equal to 1:

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

Now, we convert this equation to the standard form by completing the square. Rearrange the terms:

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

Complete the square for the x-terms:

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

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

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

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

Comparing this with the standard equation, we get:

h = \frac{1}{4}

k = 0

r = \frac{1}{4}

Final Answer: The centre of the circle is

\left(\frac{1}{4}, 0\right)

and the radius is

\frac{1}{4}

.

Q10Exercise 10.1

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

Solution

Given: The circle passes through points A(4, 1) and B(6, 5). The centre of the circle, C(h, k), lies on the line

4x + y = 16

.

To Find: The equation of the circle.

Let: Let the equation of the circle be

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

.

Solution: Since the centre (h, k) lies on the line

4x + y = 16

, we have:

4h + k = 16 \implies k = 16 - 4h \quad ...(1)

Since the circle passes through points A(4, 1) and B(6, 5), the distance from the centre C(h, k) to A and B must be equal to the radius

r

. Therefore,

CA^2 = CB^2 = r^2

.

CA^2 = (4 - h)^2 + (1 - k)^2

CB^2 = (6 - h)^2 + (5 - k)^2

Setting

CA^2 = CB^2

:

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

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

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

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

4h + 8k = 44

h + 2k = 11 \quad ...(2)

Now we solve the system of linear equations (1) and (2). Substitute

k = 16 - 4h

from (1) into (2):

h + 2(16 - 4h) = 11

h + 32 - 8h = 11

-7h = 11 - 32

-7h = -21

h = 3

Now find

k

using equation (1):

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

So, the centre of the circle is

(h, k) = (3, 4)

. Now we find the radius squared,

r^2

, using point A(4, 1):

r^2 = (4 - h)^2 + (1 - k)^2 = (4 - 3)^2 + (1 - 4)^2

r^2 = 1^2 + (-3)^2 = 1 + 9 = 10

The equation of the circle is

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

.

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

Expanding this gives:

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

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

Final Answer: The equation of the circle is

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

or

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

.

Q11Exercise 10.1

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

Solution

Given: The circle passes through points A(2, 3) and B(-1, 1). The centre of the circle, C(h, k), lies on the line

x - 3y - 11 = 0

.

To Find: The equation of the circle.

Let: Let the equation of the circle be

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

.

Solution: Since the centre (h, k) lies on the line

x - 3y - 11 = 0

, we have:

h - 3k - 11 = 0 \implies h = 3k + 11 \quad ...(1)

Since the circle passes through points A(2, 3) and B(-1, 1), the distance from the centre C(h, k) to A and B must be equal to the radius

r

. Therefore,

CA^2 = CB^2 = r^2

.

CA^2 = (2 - h)^2 + (3 - k)^2

CB^2 = (-1 - h)^2 + (1 - k)^2

Setting

CA^2 = CB^2

:

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

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

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

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

11 = 6h + 4k \quad ...(2)

Now we solve the system of linear equations (1) and (2). Substitute

h = 3k + 11

from (1) into (2):

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

11 = 18k + 66 + 4k

11 - 66 = 22k

-55 = 22k

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

Now find

h

using equation (1):

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

So, the centre of the circle is

(h, k) = \left(\frac{7}{2}, -\frac{5}{2}\right)

. Now we find the radius squared,

r^2

, using point A(2, 3):

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

r^2 = \left(\frac{4-7}{2}\right)^2 + \left(\frac{6+5}{2}\right)^2 = \left(-\frac{3}{2}\right)^2 + \left(\frac{11}{2}\right)^2

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

The equation of the circle is

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

.

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

Expanding this gives:

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

x^2 + y^2 - 7x + 5y + \frac{74}{4} = \frac{130}{4}

x^2 + y^2 - 7x + 5y = \frac{56}{4}

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

Final Answer: The equation of the circle is

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

or

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

.

Q12Exercise 10.1

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

Solution

Given: Radius of the circle

r = 5

. The centre lies on the x-axis. The circle passes through the point P(2, 3).

To Find: The equation of the circle.

Let: Since the centre lies on the x-axis, its coordinates are of the form C(h, 0).

Solution: The equation of a circle with centre (h, 0) and radius 5 is:

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

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

Since the circle passes through the point (2, 3), these coordinates must satisfy the equation:

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

(2 - h)^2 + 9 = 25

(2 - h)^2 = 16

Taking the square root of both sides:

2 - h = \pm 4

This gives two possible values for

h

: Case 1:

2 - h = 4 \implies h = 2 - 4 = -2

Case 2:

2 - h = -4 \implies h = 2 + 4 = 6

So, there are two possible circles that satisfy the given conditions.

For Case 1, the centre is

(-2, 0)

. The equation is:

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

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

For Case 2, the centre is

(6, 0)

. The equation is:

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

Final Answer: There are two possible equations for the circle:

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

and

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

.

Q13Exercise 10.1

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

Solution

Given: The circle passes through the origin (0, 0). The circle makes an intercept of length

a

on the x-axis and an intercept of length

b

on the y-axis.

To Find: The equation of the circle.

Solution: Since the circle passes through the origin (0, 0) and makes intercepts

a

and

b

on the coordinate axes, it must also pass through the points A(a, 0) and B(0, b). Let the equation of the circle be

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

. Since the points (0, 0), (a, 0), and (0, b) lie on the circle, they must satisfy the equation. For (0, 0):

(0 - h)^2 + (0 - k)^2 = r^2 \implies h^2 + k^2 = r^2 \quad ...(1)

For (a, 0):

(a - h)^2 + (0 - k)^2 = r^2 \implies a^2 - 2ah + h^2 + k^2 = r^2 \quad ...(2)

For (0, b):

(0 - h)^2 + (b - k)^2 = r^2 \implies h^2 + b^2 - 2bk + k^2 = r^2 \quad ...(3)

Substitute

r^2 = h^2 + k^2

from (1) into (2):

a^2 - 2ah + h^2 + k^2 = h^2 + k^2

a^2 - 2ah = 0

a(a - 2h) = 0

Since the intercept

a

is non-zero, we have

a - 2h = 0

, which means

h = \frac{a}{2}

.

Substitute

r^2 = h^2 + k^2

from (1) into (3):

h^2 + b^2 - 2bk + k^2 = h^2 + k^2

b^2 - 2bk = 0

b(b - 2k) = 0

Since the intercept

b

is non-zero, we have

b - 2k = 0

, which means

k = \frac{b}{2}

.

So, the centre of the circle is

(h, k) = \left(\frac{a}{2}, \frac{b}{2}\right)

. Now, find the radius squared using equation (1):

r^2 = h^2 + k^2 = \left(\frac{a}{2}\right)^2 + \left(\frac{b}{2}\right)^2 = \frac{a^2}{4} + \frac{b^2}{4}

The equation of the circle is:

\left(x - \frac{a}{2}\right)^2 + \left(y - \frac{b}{2}\right)^2 = \frac{a^2 + b^2}{4}

Expanding this equation:

x^2 - ax + \frac{a^2}{4} + y^2 - by + \frac{b^2}{4} = \frac{a^2}{4} + \frac{b^2}{4}

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

Final Answer: The equation of the circle is

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

.

Q14Exercise 10.1

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

Solution

Given: Centre of the circle

(h, k) = (2, 2)

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

To Find: The equation of the circle.

Solution: The radius

r

of the circle is the distance between the centre C(2, 2) and the point P(4, 5) on the circle. Using the distance formula,

r^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2

.

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

r^2 = 2^2 + 3^2

r^2 = 4 + 9 = 13

Now we have the centre

(h, k) = (2, 2)

and the radius squared

r^2 = 13

. The equation of the circle is given by the formula

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

. Substituting the values:

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

Expanding the equation:

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

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

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

Final Answer: The equation of the circle is

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

or

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

.

Q15Exercise 10.1

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

Solution

Given: The equation of the circle is

x^2 + y^2 = 25

. The point to check is P(-2.5, 3.5).

To Find: Whether the point lies inside, outside, or on the circle.

Solution: The given circle equation

x^2 + y^2 = 25

has its centre at the origin C(0, 0) and its radius squared is

r^2 = 25

, so the radius is

r = 5

. To determine the position of the point P(-2.5, 3.5) with respect to the circle, we calculate the square of the distance from the centre C to the point P, let's call it

d^2

.

d^2 = (-2.5 - 0)^2 + (3.5 - 0)^2

d^2 = (-2.5)^2 + (3.5)^2

d^2 = 6.25 + 12.25

d^2 = 18.5

Now, we compare

d^2

with

r^2

.

d^2 = 18.5

and

r^2 = 25

. Since

d^2 < r^2

(

18.5 < 25

), the distance from the centre to the point is less than the radius. Therefore, the point lies inside the circle.

Final Answer: The point

(-2.5, 3.5)

lies inside the circle

x^2 + y^2 = 25

.

Q1Exercise 10.2

In each of the following Exercises 1 to 6, find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum. $y^{2}=12 x$

Solution

Given: The equation of the parabola is

y^2 = 12x

.

To Find: Coordinates of the focus, the axis, the equation of the directrix, and the length of the latus rectum.

Solution: The given equation is of the form

y^2 = 4ax

. This is a parabola opening to the right with its vertex at the origin (0,0) and axis along the x-axis. Comparing

y^2 = 12x

with

y^2 = 4ax

, we get:

4a = 12 \implies a = 3

Now we can find the required properties:

Focus: The coordinates of the focus are $(a, 0)$ . Focus = $(3, 0)$ .
Axis: The axis of symmetry is the x-axis. Equation of the axis is $y = 0$ .
Directrix: The equation of the directrix is $x = -a$ . Directrix: $x = -3$ or $x + 3 = 0$ .
Latus Rectum: The length of the latus rectum is $4a$ . Length of latus rectum = $12$ .

Final Answer:

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

Q2Exercise 10.2

$x^{2}=6 y$

Solution

Given: The equation of the parabola is

x^2 = 6y

.

To Find: Coordinates of the focus, the axis, the equation of the directrix, and the length of the latus rectum.

Solution: The given equation is of the form

x^2 = 4ay

. This is a parabola opening upwards with its vertex at the origin (0,0) and axis along the y-axis. Comparing

x^2 = 6y

with

x^2 = 4ay

, we get:

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

Now we can find the required properties:

Focus: The coordinates of the focus are $(0, a)$ . Focus = $\left(0, \frac{3}{2}\right)$ .
Axis: The axis of symmetry is the y-axis. Equation of the axis is $x = 0$ .
Directrix: The equation of the directrix is $y = -a$ . Directrix: $y = -\frac{3}{2}$ or $2y + 3 = 0$ .
Latus Rectum: The length of the latus rectum is $4a$ . Length of latus rectum = $6$ .

Final Answer:

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

Q3Exercise 10.2

$y^{2}=-8 x$

Solution

Given: The equation of the parabola is

y^2 = -8x

.

To Find: Coordinates of the focus, the axis, the equation of the directrix, and the length of the latus rectum.

Solution: The given equation is of the form

y^2 = -4ax

. This is a parabola opening to the left with its vertex at the origin (0,0) and axis along the x-axis. Comparing

y^2 = -8x

with

y^2 = -4ax

, we get:

4a = 8 \implies a = 2

Now we can find the required properties:

Focus: The coordinates of the focus are $(-a, 0)$ . Focus = $(-2, 0)$ .
Axis: The axis of symmetry is the x-axis. Equation of the axis is $y = 0$ .
Directrix: The equation of the directrix is $x = a$ . Directrix: $x = 2$ or $x - 2 = 0$ .
Latus Rectum: The length of the latus rectum is $4a$ . Length of latus rectum = $8$ .

Final Answer:

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

Q4Exercise 10.2

$x^{2}=-16 y$

Solution

Given: The equation of the parabola is

x^2 = -16y

.

To Find: Coordinates of the focus, the axis, the equation of the directrix, and the length of the latus rectum.

Solution: The given equation is of the form

x^2 = -4ay

. This is a parabola opening downwards with its vertex at the origin (0,0) and axis along the y-axis. Comparing

x^2 = -16y

with

x^2 = -4ay

, we get:

4a = 16 \implies a = 4

Now we can find the required properties:

Focus: The coordinates of the focus are $(0, -a)$ . Focus = $(0, -4)$ .
Axis: The axis of symmetry is the y-axis. Equation of the axis is $x = 0$ .
Directrix: The equation of the directrix is $y = a$ . Directrix: $y = 4$ or $y - 4 = 0$ .
Latus Rectum: The length of the latus rectum is $4a$ . Length of latus rectum = $16$ .

Final Answer:

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

Q5Exercise 10.2

$y^{2}=10 x$

Solution

Given: The equation of the parabola is

y^2 = 10x

.

To Find: Coordinates of the focus, the axis, the equation of the directrix, and the length of the latus rectum.

Solution: The given equation is of the form

y^2 = 4ax

. This is a parabola opening to the right with its vertex at the origin (0,0) and axis along the x-axis. Comparing

y^2 = 10x

with

y^2 = 4ax

, we get:

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

Now we can find the required properties:

Focus: The coordinates of the focus are $(a, 0)$ . Focus = $\left(\frac{5}{2}, 0\right)$ .
Axis: The axis of symmetry is the x-axis. Equation of the axis is $y = 0$ .
Directrix: The equation of the directrix is $x = -a$ . Directrix: $x = -\frac{5}{2}$ or $2x + 5 = 0$ .
Latus Rectum: The length of the latus rectum is $4a$ . Length of latus rectum = $10$ .

Final Answer:

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

Q6Exercise 10.2

$x^{2}=-9 y$

Solution

Given: The equation of the parabola is

x^2 = -9y

.

To Find: Coordinates of the focus, the axis, the equation of the directrix, and the length of the latus rectum.

Solution: The given equation is of the form

x^2 = -4ay

. This is a parabola opening downwards with its vertex at the origin (0,0) and axis along the y-axis. Comparing

x^2 = -9y

with

x^2 = -4ay

, we get:

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

Now we can find the required properties:

Focus: The coordinates of the focus are $(0, -a)$ . Focus = $\left(0, -\frac{9}{4}\right)$ .
Axis: The axis of symmetry is the y-axis. Equation of the axis is $x = 0$ .
Directrix: The equation of the directrix is $y = a$ . Directrix: $y = \frac{9}{4}$ or $4y - 9 = 0$ .
Latus Rectum: The length of the latus rectum is $4a$ . Length of latus rectum = $9$ .

Final Answer:

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

Q7Exercise 10.2

In each of the Exercises 7 to 12, find the equation of the parabola that satisfies the given conditions: 7. Focus $(6,0)$ ; directrix $x=-6$

Solution

Given: Focus is at F(6, 0). Directrix is the line

x = -6

.

To Find: The equation of the parabola.

Solution:

Determine the type of parabola: The focus (6, 0) lies on the x-axis. The directrix $x = -6$ is a vertical line. This means the axis of the parabola is the x-axis. Since the focus is to the right of the directrix, the parabola opens to the right.
Find the vertex: The vertex is the midpoint between the focus and the point on the directrix where the axis intersects it. The intersection point is (-6, 0). The vertex V is the midpoint of F(6,0) and (-6,0), which is V(0,0).
Find the value of 'a': The distance from the vertex to the focus is 'a'. $a = \sqrt{(6-0)^2 + (0-0)^2} = 6$ .
Write the equation: The standard equation for a parabola with vertex at the origin, opening to the right, is $y^2 = 4ax$ . Substituting $a = 6$ : $y^2 = 4(6)x$ $y^2 = 24x$

Final Answer: The equation of the parabola is

y^2 = 24x

.

Q8Exercise 10.2

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

Solution

Given: Focus is at F(0, -3). Directrix is the line

y = 3

.

To Find: The equation of the parabola.

Solution:

Determine the type of parabola: The focus (0, -3) lies on the y-axis. The directrix $y = 3$ is a horizontal line. This means the axis of the parabola is the y-axis. Since the focus is below the directrix, the parabola opens downwards.
Find the vertex: The vertex is the midpoint between the focus and the point on the directrix where the axis intersects it. The intersection point is (0, 3). The vertex V is the midpoint of F(0,-3) and (0,3), which is V(0,0).
Find the value of 'a': The distance from the vertex to the focus is 'a'. $a = \sqrt{(0-0)^2 + (-3-0)^2} = 3$ .
Write the equation: The standard equation for a parabola with vertex at the origin, opening downwards, is $x^2 = -4ay$ . Substituting $a = 3$ : $x^2 = -4(3)y$ $x^2 = -12y$

Final Answer: The equation of the parabola is

x^2 = -12y

.

Q9Exercise 10.2

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

Solution

Given: Vertex is at V(0, 0). Focus is at F(3, 0).

To Find: The equation of the parabola.

Solution:

Determine the type of parabola: The vertex is at the origin and the focus is on the positive x-axis. This means the parabola opens to the right and its axis is the x-axis.
Find the value of 'a': The distance from the vertex to the focus is 'a'. $a = \sqrt{(3-0)^2 + (0-0)^2} = 3$ .
Write the equation: The standard equation for a parabola with vertex at the origin, opening to the right, is $y^2 = 4ax$ . Substituting $a = 3$ : $y^2 = 4(3)x$ $y^2 = 12x$

Final Answer: The equation of the parabola is

y^2 = 12x

.

Q10Exercise 10.2

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

Solution

Given: Vertex is at V(0, 0). Focus is at F(-2, 0).

To Find: The equation of the parabola.

Solution:

Determine the type of parabola: The vertex is at the origin and the focus is on the negative x-axis. This means the parabola opens to the left and its axis is the x-axis.
Find the value of 'a': The distance from the vertex to the focus is 'a'. $a = \sqrt{(-2-0)^2 + (0-0)^2} = 2$ .
Write the equation: The standard equation for a parabola with vertex at the origin, opening to the left, is $y^2 = -4ax$ . Substituting $a = 2$ : $y^2 = -4(2)x$ $y^2 = -8x$

Final Answer: The equation of the parabola is

y^2 = -8x

.

Q11Exercise 10.2

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

Solution

Given: Vertex is at V(0, 0). Axis is along the x-axis. The parabola passes through the point P(2, 3).

To Find: The equation of the parabola.

Solution:

Determine the type of parabola: Since the vertex is at the origin and the axis is along the x-axis, the equation is either $y^2 = 4ax$ (opens right) or $y^2 = -4ax$ (opens left).
Use the given point: The parabola passes through the point (2, 3). Since the x-coordinate is positive, the parabola must open to the right. Therefore, the equation is of the form $y^2 = 4ax$ .
Find the value of 'a': Substitute the coordinates of the point (2, 3) into the equation to find 'a'. $3^2 = 4a(2)$ $9 = 8a$ $a = \frac{9}{8}$
Write the equation: Substitute the value of 'a' back into the standard equation. $y^2 = 4\left(\frac{9}{8}\right)x$ $y^2 = \frac{9}{2}x$ $2y^2 = 9x$

Final Answer: The equation of the parabola is

2y^2 = 9x

.

Q12Exercise 10.2

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

Solution

Given: Vertex is at V(0, 0). The parabola is symmetric with respect to the y-axis. The parabola passes through the point P(5, 2).

To Find: The equation of the parabola.

Solution:

Determine the type of parabola: Since the vertex is at the origin and the axis of symmetry is the y-axis, the equation is either $x^2 = 4ay$ (opens upwards) or $x^2 = -4ay$ (opens downwards).
Use the given point: The parabola passes through the point (5, 2). Since the y-coordinate is positive, the parabola must open upwards. Therefore, the equation is of the form $x^2 = 4ay$ .
Find the value of 'a': Substitute the coordinates of the point (5, 2) into the equation to find 'a'. $5^2 = 4a(2)$ $25 = 8a$ $a = \frac{25}{8}$
Write the equation: Substitute the value of 'a' back into the standard equation. $x^2 = 4\left(\frac{25}{8}\right)y$ $x^2 = \frac{25}{2}y$ $2x^2 = 25y$

Final Answer: The equation of the parabola is

2x^2 = 25y

.

Q1Exercise 10.3

In each of the Exercises 1 to 9, find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse. $\frac{x^{2}}{36}+\frac{y^{2}}{16}=1$

Solution

Given: The equation of the ellipse is

\frac{x^2}{36} + \frac{y^2}{16} = 1

.

To Find: Coordinates of the foci, vertices, lengths of major and minor axes, eccentricity, and length of the latus rectum.

Solution: The equation is of the form

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

, where

a^2 > b^2

. The major axis is along the x-axis. Comparing the given equation, we have:

a^2 = 36 \implies a = 6

b^2 = 16 \implies b = 4

We find

c

using the relation

c^2 = a^2 - b^2

:

c^2 = 36 - 16 = 20 \implies c = \sqrt{20} = 2\sqrt{5}

Now we can find the required properties:

Foci: The coordinates are $(\pm c, 0)$ . Foci: $(\pm 2\sqrt{5}, 0)$ .
Vertices: The coordinates are $(\pm a, 0)$ . Vertices: $(\pm 6, 0)$ .
Length of Major Axis: $2a = 2(6) = 12$ .
Length of Minor Axis: $2b = 2(4) = 8$ .
Eccentricity: $e = \frac{c}{a} = \frac{2\sqrt{5}}{6} = \frac{\sqrt{5}}{3}$ .
Length of Latus Rectum: $\frac{2b^2}{a} = \frac{2(16)}{6} = \frac{32}{6} = \frac{16}{3}$ .

Final Answer:

Foci: $(\pm 2\sqrt{5}, 0)$
Vertices: $(\pm 6, 0)$
Length of Major Axis: 12
Length of Minor Axis: 8
Eccentricity: $\frac{\sqrt{5}}{3}$
Length of Latus Rectum: $\frac{16}{3}$

Q2Exercise 10.3

$\frac{x^{2}}{4}+\frac{y^{2}}{25}=1$

Solution

Given: The equation of the ellipse is

\frac{x^2}{4} + \frac{y^2}{25} = 1

.

To Find: Coordinates of the foci, vertices, lengths of major and minor axes, eccentricity, and length of the latus rectum.

Solution: The equation is of the form

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

, where

a^2 > b^2

. The major axis is along the y-axis. Comparing the given equation, we have:

a^2 = 25 \implies a = 5

b^2 = 4 \implies b = 2

We find

c

using the relation

c^2 = a^2 - b^2

:

c^2 = 25 - 4 = 21 \implies c = \sqrt{21}

Now we can find the required properties:

Foci: The coordinates are $(0, \pm c)$ . Foci: $(0, \pm \sqrt{21})$ .
Vertices: The coordinates are $(0, \pm a)$ . Vertices: $(0, \pm 5)$ .
Length of Major Axis: $2a = 2(5) = 10$ .
Length of Minor Axis: $2b = 2(2) = 4$ .
Eccentricity: $e = \frac{c}{a} = \frac{\sqrt{21}}{5}$ .
Length of Latus Rectum: $\frac{2b^2}{a} = \frac{2(4)}{5} = \frac{8}{5}$ .

Final Answer:

Foci: $(0, \pm \sqrt{21})$
Vertices: $(0, \pm 5)$
Length of Major Axis: 10
Length of Minor Axis: 4
Eccentricity: $\frac{\sqrt{21}}{5}$
Length of Latus Rectum: $\frac{8}{5}$

Q3Exercise 10.3

$\frac{x^{2}}{16}+\frac{y^{2}}{9}=1$

Solution

Given: The equation of the ellipse is

\frac{x^2}{16} + \frac{y^2}{9} = 1

.

To Find: Coordinates of the foci, vertices, lengths of major and minor axes, eccentricity, and length of the latus rectum.

Solution: The equation is of the form

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

, where

a^2 > b^2

. The major axis is along the x-axis. Comparing the given equation, we have:

a^2 = 16 \implies a = 4

b^2 = 9 \implies b = 3

We find

c

using the relation

c^2 = a^2 - b^2

:

c^2 = 16 - 9 = 7 \implies c = \sqrt{7}

Now we can find the required properties:

Foci: The coordinates are $(\pm c, 0)$ . Foci: $(\pm \sqrt{7}, 0)$ .
Vertices: The coordinates are $(\pm a, 0)$ . Vertices: $(\pm 4, 0)$ .
Length of Major Axis: $2a = 2(4) = 8$ .
Length of Minor Axis: $2b = 2(3) = 6$ .
Eccentricity: $e = \frac{c}{a} = \frac{\sqrt{7}}{4}$ .
Length of Latus Rectum: $\frac{2b^2}{a} = \frac{2(9)}{4} = \frac{18}{4} = \frac{9}{2}$ .

Final Answer:

Foci: $(\pm \sqrt{7}, 0)$
Vertices: $(\pm 4, 0)$
Length of Major Axis: 8
Length of Minor Axis: 6
Eccentricity: $\frac{\sqrt{7}}{4}$
Length of Latus Rectum: $\frac{9}{2}$

Q4Exercise 10.3

$\frac{x^{2}}{25}+\frac{y^{2}}{100}=1$

Solution

Given: The equation of the ellipse is

\frac{x^2}{25} + \frac{y^2}{100} = 1

.

To Find: Coordinates of the foci, vertices, lengths of major and minor axes, eccentricity, and length of the latus rectum.

Solution: The equation is of the form

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

, where

a^2 > b^2

. The major axis is along the y-axis. Comparing the given equation, we have:

a^2 = 100 \implies a = 10

b^2 = 25 \implies b = 5

We find

c

using the relation

c^2 = a^2 - b^2

:

c^2 = 100 - 25 = 75 \implies c = \sqrt{75} = 5\sqrt{3}

Now we can find the required properties:

Foci: The coordinates are $(0, \pm c)$ . Foci: $(0, \pm 5\sqrt{3})$ .
Vertices: The coordinates are $(0, \pm a)$ . Vertices: $(0, \pm 10)$ .
Length of Major Axis: $2a = 2(10) = 20$ .
Length of Minor Axis: $2b = 2(5) = 10$ .
Eccentricity: $e = \frac{c}{a} = \frac{5\sqrt{3}}{10} = \frac{\sqrt{3}}{2}$ .
Length of Latus Rectum: $\frac{2b^2}{a} = \frac{2(25)}{10} = \frac{50}{10} = 5$ .

Final Answer:

Foci: $(0, \pm 5\sqrt{3})$
Vertices: $(0, \pm 10)$
Length of Major Axis: 20
Length of Minor Axis: 10
Eccentricity: $\frac{\sqrt{3}}{2}$
Length of Latus Rectum: 5

Q5Exercise 10.3

$\frac{x^{2}}{49}+\frac{y^{2}}{36}=1$

Solution

Given: The equation of the ellipse is

\frac{x^2}{49} + \frac{y^2}{36} = 1

.

To Find: Coordinates of the foci, vertices, lengths of major and minor axes, eccentricity, and length of the latus rectum.

Solution: The equation is of the form

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

, where

a^2 > b^2

. The major axis is along the x-axis. Comparing the given equation, we have:

a^2 = 49 \implies a = 7

b^2 = 36 \implies b = 6

We find

c

using the relation

c^2 = a^2 - b^2

:

c^2 = 49 - 36 = 13 \implies c = \sqrt{13}

Now we can find the required properties:

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

Final Answer:

Foci: $(\pm \sqrt{13}, 0)$
Vertices: $(\pm 7, 0)$
Length of Major Axis: 14
Length of Minor Axis: 12
Eccentricity: $\frac{\sqrt{13}}{7}$
Length of Latus Rectum: $\frac{72}{7}$

Q6Exercise 10.3

$\frac{x^{2}}{100}+\frac{y^{2}}{400}=1$

Solution

Given: The equation of the ellipse is

\frac{x^2}{100} + \frac{y^2}{400} = 1

.

To Find: Coordinates of the foci, vertices, lengths of major and minor axes, eccentricity, and length of the latus rectum.

Solution: The equation is of the form

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

, where

a^2 > b^2

. The major axis is along the y-axis. Comparing the given equation, we have:

a^2 = 400 \implies a = 20

b^2 = 100 \implies b = 10

We find

c

using the relation

c^2 = a^2 - b^2

:

c^2 = 400 - 100 = 300 \implies c = \sqrt{300} = 10\sqrt{3}

Now we can find the required properties:

Foci: The coordinates are $(0, \pm c)$ . Foci: $(0, \pm 10\sqrt{3})$ .
Vertices: The coordinates are $(0, \pm a)$ . Vertices: $(0, \pm 20)$ .
Length of Major Axis: $2a = 2(20) = 40$ .
Length of Minor Axis: $2b = 2(10) = 20$ .
Eccentricity: $e = \frac{c}{a} = \frac{10\sqrt{3}}{20} = \frac{\sqrt{3}}{2}$ .
Length of Latus Rectum: $\frac{2b^2}{a} = \frac{2(100)}{20} = \frac{200}{20} = 10$ .

Final Answer:

Foci: $(0, \pm 10\sqrt{3})$
Vertices: $(0, \pm 20)$
Length of Major Axis: 40
Length of Minor Axis: 20
Eccentricity: $\frac{\sqrt{3}}{2}$
Length of Latus Rectum: 10

Q7Exercise 10.3

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

Solution

Given: The equation of the ellipse is

36x^2 + 4y^2 = 144

.

To Find: Coordinates of the foci, vertices, lengths of major and minor axes, eccentricity, and length of the latus rectum.

Solution: First, we write the equation in standard form by dividing by 144:

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

\frac{x^2}{4} + \frac{y^2}{36} = 1

The equation is of the form

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

, where

a^2 > b^2

. The major axis is along the y-axis. Comparing the equation, we have:

a^2 = 36 \implies a = 6

b^2 = 4 \implies b = 2

We find

c

using the relation

c^2 = a^2 - b^2

:

c^2 = 36 - 4 = 32 \implies c = \sqrt{32} = 4\sqrt{2}

Now we can find the required properties:

Foci: The coordinates are $(0, \pm c)$ . Foci: $(0, \pm 4\sqrt{2})$ .
Vertices: The coordinates are $(0, \pm a)$ . Vertices: $(0, \pm 6)$ .
Length of Major Axis: $2a = 2(6) = 12$ .
Length of Minor Axis: $2b = 2(2) = 4$ .
Eccentricity: $e = \frac{c}{a} = \frac{4\sqrt{2}}{6} = \frac{2\sqrt{2}}{3}$ .
Length of Latus Rectum: $\frac{2b^2}{a} = \frac{2(4)}{6} = \frac{8}{6} = \frac{4}{3}$ .

Final Answer:

Foci: $(0, \pm 4\sqrt{2})$
Vertices: $(0, \pm 6)$
Length of Major Axis: 12
Length of Minor Axis: 4
Eccentricity: $\frac{2\sqrt{2}}{3}$
Length of Latus Rectum: $\frac{4}{3}$

Q8Exercise 10.3

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

Solution

Given: The equation of the ellipse is

16x^2 + y^2 = 16

.

To Find: Coordinates of the foci, vertices, lengths of major and minor axes, eccentricity, and length of the latus rectum.

Solution: First, we write the equation in standard form by dividing by 16:

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

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

The equation is of the form

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

, where

a^2 > b^2

. The major axis is along the y-axis. Comparing the equation, we have:

a^2 = 16 \implies a = 4

b^2 = 1 \implies b = 1

We find

c

using the relation

c^2 = a^2 - b^2

:

c^2 = 16 - 1 = 15 \implies c = \sqrt{15}

Now we can find the required properties:

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

Final Answer:

Foci: $(0, \pm \sqrt{15})$
Vertices: $(0, \pm 4)$
Length of Major Axis: 8
Length of Minor Axis: 2
Eccentricity: $\frac{\sqrt{15}}{4}$
Length of Latus Rectum: $\frac{1}{2}$

Q9Exercise 10.3

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

Solution

Given: The equation of the ellipse is

4x^2 + 9y^2 = 36

.

To Find: Coordinates of the foci, vertices, lengths of major and minor axes, eccentricity, and length of the latus rectum.

Solution: First, we write the equation in standard form by dividing by 36:

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

\frac{x^2}{9} + \frac{y^2}{4} = 1

The equation is of the form

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

, where

a^2 > b^2

. The major axis is along the x-axis. Comparing the equation, we have:

a^2 = 9 \implies a = 3

b^2 = 4 \implies b = 2

We find

c

using the relation

c^2 = a^2 - b^2

:

c^2 = 9 - 4 = 5 \implies c = \sqrt{5}

Now we can find the required properties:

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

Final Answer:

Foci: $(\pm \sqrt{5}, 0)$
Vertices: $(\pm 3, 0)$
Length of Major Axis: 6
Length of Minor Axis: 4
Eccentricity: $\frac{\sqrt{5}}{3}$
Length of Latus Rectum: $\frac{8}{3}$

Q10Exercise 10.3

In each of the following Exercises 10 to 20, find the equation for the ellipse that satisfies the given conditions: 10. Vertices $( \pm 5,0)$ , foci $( \pm 4,0)$

Solution

Given: Vertices are at

(\pm 5, 0)

. Foci are at

(\pm 4, 0)

.

To Find: The equation of the ellipse.

Solution: Since the vertices and foci are on the x-axis, the major axis is along the x-axis. The standard equation is

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

. From the vertices

(\pm a, 0)

, we have

a = 5

. From the foci

(\pm c, 0)

, we have

c = 4

. We know the relation

c^2 = a^2 - b^2

for an ellipse.

b^2 = a^2 - c^2

b^2 = 5^2 - 4^2 = 25 - 16 = 9

So,

a^2 = 5^2 = 25

and

b^2 = 9

. Substituting these values into the standard equation:

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

Final Answer: The equation of the ellipse is

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

.

Q11Exercise 10.3

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

Solution

Given: Vertices are at

(0, \pm 13)

. Foci are at

(0, \pm 5)

.

To Find: The equation of the ellipse.

Solution: Since the vertices and foci are on the y-axis, the major axis is along the y-axis. The standard equation is

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

. From the vertices

(0, \pm a)

, we have

a = 13

. From the foci

(0, \pm c)

, we have

c = 5

. We know the relation

c^2 = a^2 - b^2

for an ellipse.

b^2 = a^2 - c^2

b^2 = 13^2 - 5^2 = 169 - 25 = 144

So,

a^2 = 13^2 = 169

and

b^2 = 144

. Substituting these values into the standard equation:

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

Final Answer: The equation of the ellipse is

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

.

Q12Exercise 10.3

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

Solution

Given: Vertices are at

(\pm 6, 0)

. Foci are at

(\pm 4, 0)

.

To Find: The equation of the ellipse.

Solution: Since the vertices and foci are on the x-axis, the major axis is along the x-axis. The standard equation is

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

. From the vertices

(\pm a, 0)

, we have

a = 6

. From the foci

(\pm c, 0)

, we have

c = 4

. We know the relation

c^2 = a^2 - b^2

for an ellipse.

b^2 = a^2 - c^2

b^2 = 6^2 - 4^2 = 36 - 16 = 20

So,

a^2 = 6^2 = 36

and

b^2 = 20

. Substituting these values into the standard equation:

\frac{x^2}{36} + \frac{y^2}{20} = 1

Final Answer: The equation of the ellipse is

\frac{x^2}{36} + \frac{y^2}{20} = 1

.

Q13Exercise 10.3

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

Solution

Given: Ends of the major axis are

(\pm 3, 0)

. Ends of the minor axis are

(0, \pm 2)

.

To Find: The equation of the ellipse.

Solution: The ends of the major axis are the vertices. Since they are on the x-axis, the major axis is along the x-axis. The standard equation is

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

. From the ends of the major axis

(\pm a, 0)

, we have

a = 3

. From the ends of the minor axis

(0, \pm b)

, we have

b = 2

. So,

a^2 = 3^2 = 9

and

b^2 = 2^2 = 4

. Substituting these values into the standard equation:

\frac{x^2}{9} + \frac{y^2}{4} = 1

Final Answer: The equation of the ellipse is

\frac{x^2}{9} + \frac{y^2}{4} = 1

.

Q14Exercise 10.3

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

Solution

Given: Ends of the major axis are

(0, \pm \sqrt{5})

. Ends of the minor axis are

(\pm 1, 0)

.

To Find: The equation of the ellipse.

Solution: The ends of the major axis are the vertices. Since they are on the y-axis, the major axis is along the y-axis. The standard equation is

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

. From the ends of the major axis

(0, \pm a)

, we have

a = \sqrt{5}

. From the ends of the minor axis

(\pm b, 0)

, we have

b = 1

. So,

a^2 = (\sqrt{5})^2 = 5

and

b^2 = 1^2 = 1

. Substituting these values into the standard equation:

\frac{x^2}{1} + \frac{y^2}{5} = 1

Final Answer: The equation of the ellipse is

x^2 + \frac{y^2}{5} = 1

.

Q15Exercise 10.3

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

Solution

Given: Length of the major axis is 26. Foci are at

(\pm 5, 0)

.

To Find: The equation of the ellipse.

Solution: Since the foci are on the x-axis, the major axis is along the x-axis. The standard equation is

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

. Length of the major axis is

2a = 26

, so

a = 13

. From the foci

(\pm c, 0)

, we have

c = 5

. We know the relation

c^2 = a^2 - b^2

for an ellipse.

b^2 = a^2 - c^2

b^2 = 13^2 - 5^2 = 169 - 25 = 144

So,

a^2 = 13^2 = 169

and

b^2 = 144

. Substituting these values into the standard equation:

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

Final Answer: The equation of the ellipse is

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

.

Q16Exercise 10.3

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

Solution

Given: Length of the minor axis is 16. Foci are at

(0, \pm 6)

.

To Find: The equation of the ellipse.

Solution: Since the foci are on the y-axis, the major axis is along the y-axis. The standard equation is

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

. Length of the minor axis is

2b = 16

, so

b = 8

. From the foci

(0, \pm c)

, we have

c = 6

. We know the relation

c^2 = a^2 - b^2

for an ellipse.

a^2 = b^2 + c^2

a^2 = 8^2 + 6^2 = 64 + 36 = 100

So,

a^2 = 100

and

b^2 = 8^2 = 64

. Substituting these values into the standard equation:

\frac{x^2}{64} + \frac{y^2}{100} = 1

Final Answer: The equation of the ellipse is

\frac{x^2}{64} + \frac{y^2}{100} = 1

.

Q17Exercise 10.3

Foci $( \pm 3,0), a=4$

Solution

Given: Foci are at

(\pm 3, 0)

. Semi-major axis

a = 4

.

To Find: The equation of the ellipse.

Solution: Since the foci are on the x-axis, the major axis is along the x-axis. The standard equation is

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

. From the foci

(\pm c, 0)

, we have

c = 3

. We are given

a = 4

. We know the relation

c^2 = a^2 - b^2

for an ellipse.

b^2 = a^2 - c^2

b^2 = 4^2 - 3^2 = 16 - 9 = 7

So,

a^2 = 4^2 = 16

and

b^2 = 7

. Substituting these values into the standard equation:

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

Final Answer: The equation of the ellipse is

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

.

Q18Exercise 10.3

$b=3, c=4$ , centre at the origin; foci on the $x$ axis.

Solution

Given: Semi-minor axis

b = 3

. Distance from centre to focus

c = 4

. Centre is at the origin. Foci are on the x-axis.

To Find: The equation of the ellipse.

Solution: Since the foci are on the x-axis, the major axis is along the x-axis. The standard equation is

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

. We are given

b = 3

and

c = 4

. We know the relation

c^2 = a^2 - b^2

for an ellipse.

a^2 = b^2 + c^2

a^2 = 3^2 + 4^2 = 9 + 16 = 25

So,

a^2 = 25

and

b^2 = 3^2 = 9

. Substituting these values into the standard equation:

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

Final Answer: The equation of the ellipse is

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

.

Q19Exercise 10.3

Centre at $(0,0)$ , major axis on the $y$ -axis and passes through the points $(3,2)$ and $(1,6)$ .

Solution

Given: Centre is at (0, 0). Major axis is on the y-axis. The ellipse passes through points P(3, 2) and Q(1, 6).

To Find: The equation of the ellipse.

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

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

. Since the ellipse passes through (3, 2), we have:

\frac{3^2}{b^2} + \frac{2^2}{a^2} = 1 \implies \frac{9}{b^2} + \frac{4}{a^2} = 1 \quad ...(1)

Since the ellipse passes through (1, 6), we have:

\frac{1^2}{b^2} + \frac{6^2}{a^2} = 1 \implies \frac{1}{b^2} + \frac{36}{a^2} = 1 \quad ...(2)

Let

u = \frac{1}{b^2}

and

v = \frac{1}{a^2}

. The equations become:

9u + 4v = 1 \quad ...(3)

u + 36v = 1 \quad ...(4)

From equation (4),

u = 1 - 36v

. Substitute this into equation (3):

9(1 - 36v) + 4v = 1

9 - 324v + 4v = 1

9 - 320v = 1

8 = 320v

v = \frac{8}{320} = \frac{1}{40}

Now find

u

:

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

Since

u = \frac{1}{b^2}

and

v = \frac{1}{a^2}

, we have:

b^2 = 10

and

a^2 = 40

. Substituting these values into the standard equation:

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

Final Answer: The equation of the ellipse is

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

.

Q20Exercise 10.3

Major axis on the $x$ -axis and passes through the points $(4,3)$ and $(6,2)$ .

Solution

Given: Major axis is on the x-axis. The ellipse passes through points P(4, 3) and Q(6, 2). Centre is at the origin (implied by standard form).

To Find: The equation of the ellipse.

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

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

. Since the ellipse passes through (4, 3), we have:

\frac{4^2}{a^2} + \frac{3^2}{b^2} = 1 \implies \frac{16}{a^2} + \frac{9}{b^2} = 1 \quad ...(1)

Since the ellipse passes through (6, 2), we have:

\frac{6^2}{a^2} + \frac{2^2}{b^2} = 1 \implies \frac{36}{a^2} + \frac{4}{b^2} = 1 \quad ...(2)

Let

u = \frac{1}{a^2}

and

v = \frac{1}{b^2}

. The equations become:

16u + 9v = 1 \quad ...(3)

36u + 4v = 1 \quad ...(4)

Multiply equation (3) by 4 and equation (4) by 9:

64u + 36v = 4 \quad ...(5)

324u + 36v = 9 \quad ...(6)

Subtract equation (5) from equation (6):

(324u - 64u) + (36v - 36v) = 9 - 4

260u = 5

u = \frac{5}{260} = \frac{1}{52}

Substitute

u = \frac{1}{52}

into equation (3):

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

\frac{4}{13} + 9v = 1

9v = 1 - \frac{4}{13} = \frac{9}{13}

v = \frac{1}{13}

Since

u = \frac{1}{a^2}

and

v = \frac{1}{b^2}

, we have:

a^2 = 52

and

b^2 = 13

. Substituting these values into the standard equation:

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

Final Answer: The equation of the ellipse is

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

.

Q1Exercise 10.4

In each of the Exercises 1 to 6, find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbolas. $\frac{x^{2}}{16}-\frac{y^{2}}{9}=1$

Solution

Given: The equation of the hyperbola is

\frac{x^2}{16} - \frac{y^2}{9} = 1

.

To Find: Coordinates of the foci, vertices, eccentricity, and the length of the latus rectum.

Solution: The equation is of the form

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

. The transverse axis is along the x-axis. Comparing the given equation, we have:

a^2 = 16 \implies a = 4

b^2 = 9 \implies b = 3

We find

c

using the relation

c^2 = a^2 + b^2

:

c^2 = 16 + 9 = 25 \implies c = 5

Now we can find the required properties:

Foci: The coordinates are $(\pm c, 0)$ . Foci: $(\pm 5, 0)$ .
Vertices: The coordinates are $(\pm a, 0)$ . Vertices: $(\pm 4, 0)$ .
Eccentricity: $e = \frac{c}{a} = \frac{5}{4}$ .
Length of Latus Rectum: $\frac{2b^2}{a} = \frac{2(9)}{4} = \frac{18}{4} = \frac{9}{2}$ .

Final Answer:

Foci: $(\pm 5, 0)$
Vertices: $(\pm 4, 0)$
Eccentricity: $\frac{5}{4}$
Length of Latus Rectum: $\frac{9}{2}$

Q2Exercise 10.4

$\frac{y^{2}}{9}-\frac{x^{2}}{27}=1$

Solution

Given: The equation of the hyperbola is

\frac{y^2}{9} - \frac{x^2}{27} = 1

.

To Find: Coordinates of the foci, vertices, eccentricity, and the length of the latus rectum.

Solution: The equation is of the form

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

. The transverse axis is along the y-axis. Comparing the given equation, we have:

a^2 = 9 \implies a = 3

b^2 = 27 \implies b = \sqrt{27} = 3\sqrt{3}

We find

c

using the relation

c^2 = a^2 + b^2

:

c^2 = 9 + 27 = 36 \implies c = 6

Now we can find the required properties:

Foci: The coordinates are $(0, \pm c)$ . Foci: $(0, \pm 6)$ .
Vertices: The coordinates are $(0, \pm a)$ . Vertices: $(0, \pm 3)$ .
Eccentricity: $e = \frac{c}{a} = \frac{6}{3} = 2$ .
Length of Latus Rectum: $\frac{2b^2}{a} = \frac{2(27)}{3} = 2(9) = 18$ .

Final Answer:

Foci: $(0, \pm 6)$
Vertices: $(0, \pm 3)$
Eccentricity: 2
Length of Latus Rectum: 18

Q3Exercise 10.4

$9 y^{2}-4 x^{2}=36$

Solution

Given: The equation of the hyperbola is

9y^2 - 4x^2 = 36

.

To Find: Coordinates of the foci, vertices, eccentricity, and the length of the latus rectum.

Solution: First, we write the equation in standard form by dividing by 36:

\frac{9y^2}{36} - \frac{4x^2}{36} = 1

\frac{y^2}{4} - \frac{x^2}{9} = 1

The equation is of the form

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

. The transverse axis is along the y-axis. Comparing the equation, we have:

a^2 = 4 \implies a = 2

b^2 = 9 \implies b = 3

We find

c

using the relation

c^2 = a^2 + b^2

:

c^2 = 4 + 9 = 13 \implies c = \sqrt{13}

Now we can find the required properties:

Foci: The coordinates are $(0, \pm c)$ . Foci: $(0, \pm \sqrt{13})$ .
Vertices: The coordinates are $(0, \pm a)$ . Vertices: $(0, \pm 2)$ .
Eccentricity: $e = \frac{c}{a} = \frac{\sqrt{13}}{2}$ .
Length of Latus Rectum: $\frac{2b^2}{a} = \frac{2(9)}{2} = 9$ .

Final Answer:

Foci: $(0, \pm \sqrt{13})$
Vertices: $(0, \pm 2)$
Eccentricity: $\frac{\sqrt{13}}{2}$
Length of Latus Rectum: 9

Q4Exercise 10.4

$16 x^{2}-9 y^{2}=576$

Solution

Given: The equation of the hyperbola is

16x^2 - 9y^2 = 576

.

To Find: Coordinates of the foci, vertices, eccentricity, and the length of the latus rectum.

Solution: First, we write the equation in standard form by dividing by 576:

\frac{16x^2}{576} - \frac{9y^2}{576} = 1

\frac{x^2}{36} - \frac{y^2}{64} = 1

The equation is of the form

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

. The transverse axis is along the x-axis. Comparing the equation, we have:

a^2 = 36 \implies a = 6

b^2 = 64 \implies b = 8

We find

c

using the relation

c^2 = a^2 + b^2

:

c^2 = 36 + 64 = 100 \implies c = 10

Now we can find the required properties:

Foci: The coordinates are $(\pm c, 0)$ . Foci: $(\pm 10, 0)$ .
Vertices: The coordinates are $(\pm a, 0)$ . Vertices: $(\pm 6, 0)$ .
Eccentricity: $e = \frac{c}{a} = \frac{10}{6} = \frac{5}{3}$ .
Length of Latus Rectum: $\frac{2b^2}{a} = \frac{2(64)}{6} = \frac{64}{3}$ .

Final Answer:

Foci: $(\pm 10, 0)$
Vertices: $(\pm 6, 0)$
Eccentricity: $\frac{5}{3}$
Length of Latus Rectum: $\frac{64}{3}$

Q5Exercise 10.4

$5 y^{2}-9 x^{2}=36$

Solution

Given: The equation of the hyperbola is

5y^2 - 9x^2 = 36

.

To Find: Coordinates of the foci, vertices, eccentricity, and the length of the latus rectum.

Solution: First, we write the equation in standard form by dividing by 36:

\frac{5y^2}{36} - \frac{9x^2}{36} = 1

\frac{y^2}{36/5} - \frac{x^2}{4} = 1

The equation is of the form

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

. The transverse axis is along the y-axis. Comparing the equation, we have:

a^2 = \frac{36}{5} \implies a = \frac{6}{\sqrt{5}}

b^2 = 4 \implies b = 2

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} \implies c = \sqrt{\frac{56}{5}} = \frac{2\sqrt{14}}{\sqrt{5}}

Now we can find the required properties:

Foci: The coordinates are $(0, \pm c)$ . Foci: $\left(0, \pm \frac{2\sqrt{14}}{\sqrt{5}}\right)$ or $\left(0, \pm \frac{2\sqrt{70}}{5}\right)$ .
Vertices: The coordinates are $(0, \pm a)$ . Vertices: $\left(0, \pm \frac{6}{\sqrt{5}}\right)$ or $\left(0, \pm \frac{6\sqrt{5}}{5}\right)$ .
Eccentricity: $e = \frac{c}{a} = \frac{2\sqrt{14}/\sqrt{5}}{6/\sqrt{5}} = \frac{2\sqrt{14}}{6} = \frac{\sqrt{14}}{3}$ .
Length of Latus Rectum: $\frac{2b^2}{a} = \frac{2(4)}{6/\sqrt{5}} = \frac{8\sqrt{5}}{6} = \frac{4\sqrt{5}}{3}$ .

Final Answer:

Foci: $\left(0, \pm \frac{2\sqrt{70}}{5}\right)$
Vertices: $\left(0, \pm \frac{6\sqrt{5}}{5}\right)$
Eccentricity: $\frac{\sqrt{14}}{3}$
Length of Latus Rectum: $\frac{4\sqrt{5}}{3}$

Q6Exercise 10.4

$49 y^{2}-16 x^{2}=784$ .

Solution

Given: The equation of the hyperbola is

49y^2 - 16x^2 = 784

.

To Find: Coordinates of the foci, vertices, eccentricity, and the length of the latus rectum.

Solution: First, we write the equation in standard form by dividing by 784:

\frac{49y^2}{784} - \frac{16x^2}{784} = 1

\frac{y^2}{16} - \frac{x^2}{49} = 1

The equation is of the form

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

. The transverse axis is along the y-axis. Comparing the equation, we have:

a^2 = 16 \implies a = 4

b^2 = 49 \implies b = 7

We find

c

using the relation

c^2 = a^2 + b^2

:

c^2 = 16 + 49 = 65 \implies c = \sqrt{65}

Now we can find the required properties:

Foci: The coordinates are $(0, \pm c)$ . Foci: $(0, \pm \sqrt{65})$ .
Vertices: The coordinates are $(0, \pm a)$ . Vertices: $(0, \pm 4)$ .
Eccentricity: $e = \frac{c}{a} = \frac{\sqrt{65}}{4}$ .
Length of Latus Rectum: $\frac{2b^2}{a} = \frac{2(49)}{4} = \frac{49}{2}$ .

Final Answer:

Foci: $(0, \pm \sqrt{65})$
Vertices: $(0, \pm 4)$
Eccentricity: $\frac{\sqrt{65}}{4}$
Length of Latus Rectum: $\frac{49}{2}$

Q7Exercise 10.4

In each of the Exercises 7 to 15, find the equations of the hyperbola satisfying the given conditions. 7. Vertices $( \pm 2,0)$ , foci $( \pm 3,0)$

Solution

Given: Vertices are at

(\pm 2, 0)

. Foci are at

(\pm 3, 0)

.

To Find: The equation of the hyperbola.

Solution: Since the vertices and foci are on the x-axis, the transverse axis is along the x-axis. The standard equation is

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

. From the vertices

(\pm a, 0)

, we have

a = 2

. From the foci

(\pm c, 0)

, we have

c = 3

. We know the relation

c^2 = a^2 + b^2

for a hyperbola.

b^2 = c^2 - a^2

b^2 = 3^2 - 2^2 = 9 - 4 = 5

So,

a^2 = 2^2 = 4

and

b^2 = 5

. Substituting these values into the standard equation:

\frac{x^2}{4} - \frac{y^2}{5} = 1

Final Answer: The equation of the hyperbola is

\frac{x^2}{4} - \frac{y^2}{5} = 1

.

Q8Exercise 10.4

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

Solution

Given: Vertices are at

(0, \pm 5)

. Foci are at

(0, \pm 8)

.

To Find: The equation of the hyperbola.

Solution: Since the vertices and foci are on the y-axis, the transverse axis is along the y-axis. The standard equation is

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

. From the vertices

(0, \pm a)

, we have

a = 5

. From the foci

(0, \pm c)

, we have

c = 8

. We know the relation

c^2 = a^2 + b^2

for a hyperbola.

b^2 = c^2 - a^2

b^2 = 8^2 - 5^2 = 64 - 25 = 39

So,

a^2 = 5^2 = 25

and

b^2 = 39

. Substituting these values into the standard equation:

\frac{y^2}{25} - \frac{x^2}{39} = 1

Final Answer: The equation of the hyperbola is

\frac{y^2}{25} - \frac{x^2}{39} = 1

.

Q9Exercise 10.4

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

Solution

Given: Vertices are at

(0, \pm 3)

. Foci are at

(0, \pm 5)

.

To Find: The equation of the hyperbola.

Solution: Since the vertices and foci are on the y-axis, the transverse axis is along the y-axis. The standard equation is

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

. From the vertices

(0, \pm a)

, we have

a = 3

. From the foci

(0, \pm c)

, we have

c = 5

. We know the relation

c^2 = a^2 + b^2

for a hyperbola.

b^2 = c^2 - a^2

b^2 = 5^2 - 3^2 = 25 - 9 = 16

So,

a^2 = 3^2 = 9

and

b^2 = 16

. Substituting these values into the standard equation:

\frac{y^2}{9} - \frac{x^2}{16} = 1

Final Answer: The equation of the hyperbola is

\frac{y^2}{9} - \frac{x^2}{16} = 1

.

Q10Exercise 10.4

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

Solution

Given: Foci are at

(\pm 5, 0)

. Length of the transverse axis is 8.

To Find: The equation of the hyperbola.

Solution: Since the foci are on the x-axis, the transverse axis is along the x-axis. The standard equation is

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

. From the foci

(\pm c, 0)

, we have

c = 5

. The length of the transverse axis is

2a = 8

, so

a = 4

. We know the relation

c^2 = a^2 + b^2

for a hyperbola.

b^2 = c^2 - a^2

b^2 = 5^2 - 4^2 = 25 - 16 = 9

So,

a^2 = 4^2 = 16

and

b^2 = 9

. Substituting these values into the standard equation:

\frac{x^2}{16} - \frac{y^2}{9} = 1

Final Answer: The equation of the hyperbola is

\frac{x^2}{16} - \frac{y^2}{9} = 1

.

Q11Exercise 10.4

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

Solution

Given: Foci are at

(0, \pm 13)

. Length of the conjugate axis is 24.

To Find: The equation of the hyperbola.

Solution: Since the foci are on the y-axis, the transverse axis is along the y-axis. The standard equation is

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

. From the foci

(0, \pm c)

, we have

c = 13

. The length of the conjugate axis is

2b = 24

, so

b = 12

. We know the relation

c^2 = a^2 + b^2

for a hyperbola.

a^2 = c^2 - b^2

a^2 = 13^2 - 12^2 = 169 - 144 = 25

So,

a^2 = 25

and

b^2 = 12^2 = 144

. Substituting these values into the standard equation:

\frac{y^2}{25} - \frac{x^2}{144} = 1

Final Answer: The equation of the hyperbola is

\frac{y^2}{25} - \frac{x^2}{144} = 1

.

Q12Exercise 10.4

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

Solution

Given: Foci are at

(\pm 3\sqrt{5}, 0)

. Length of the latus rectum is 8.

To Find: The equation of the hyperbola.

Solution: Since the foci are on the x-axis, the transverse axis is along the x-axis. The standard equation is

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

. From the foci

(\pm c, 0)

, we have

c = 3\sqrt{5}

. The length of the latus rectum is

\frac{2b^2}{a} = 8

, which implies

b^2 = 4a

. We know the relation

c^2 = a^2 + b^2

for a hyperbola.

(3\sqrt{5})^2 = a^2 + 4a

45 = a^2 + 4a

a^2 + 4a - 45 = 0

Factorizing the quadratic equation:

(a + 9)(a - 5) = 0

Since

a

must be positive, we take

a = 5

. Now, find

b^2

:

b^2 = 4a = 4(5) = 20

So,

a^2 = 5^2 = 25

and

b^2 = 20

. Substituting these values into the standard equation:

\frac{x^2}{25} - \frac{y^2}{20} = 1

Final Answer: The equation of the hyperbola is

\frac{x^2}{25} - \frac{y^2}{20} = 1

.

Q13Exercise 10.4

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

Solution

Given: Foci are at

(\pm 4, 0)

. Length of the latus rectum is 12.

To Find: The equation of the hyperbola.

Solution: Since the foci are on the x-axis, the transverse axis is along the x-axis. The standard equation is

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

. From the foci

(\pm c, 0)

, we have

c = 4

. The length of the latus rectum is

\frac{2b^2}{a} = 12

, which implies

b^2 = 6a

. We know the relation

c^2 = a^2 + b^2

for a hyperbola.

4^2 = a^2 + 6a

16 = a^2 + 6a

a^2 + 6a - 16 = 0

Factorizing the quadratic equation:

(a + 8)(a - 2) = 0

Since

a

must be positive, we take

a = 2

. Now, find

b^2

:

b^2 = 6a = 6(2) = 12

So,

a^2 = 2^2 = 4

and

b^2 = 12

. Substituting these values into the standard equation:

\frac{x^2}{4} - \frac{y^2}{12} = 1

Final Answer: The equation of the hyperbola is

\frac{x^2}{4} - \frac{y^2}{12} = 1

.

Q14Exercise 10.4

vertices $( \pm 7,0), e=\frac{4}{3}$ .

Solution

Given: Vertices are at

(\pm 7, 0)

. Eccentricity

e = \frac{4}{3}

.

To Find: The equation of the hyperbola.

Solution: Since the vertices are on the x-axis, the transverse axis is along the x-axis. The standard equation is

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

. From the vertices

(\pm a, 0)

, we have

a = 7

. We are given the eccentricity

e = \frac{c}{a}

.

\frac{4}{3} = \frac{c}{7} \implies c = \frac{28}{3}

We know the relation

c^2 = a^2 + b^2

for a hyperbola.

b^2 = c^2 - a^2

b^2 = \left(\frac{28}{3}\right)^2 - 7^2 = \frac{784}{9} - 49 = \frac{784 - 441}{9} = \frac{343}{9}

So,

a^2 = 7^2 = 49

and

b^2 = \frac{343}{9}

. Substituting these values into the standard equation:

\frac{x^2}{49} - \frac{y^2}{343/9} = 1

\frac{x^2}{49} - \frac{9y^2}{343} = 1

Final Answer: The equation of the hyperbola is

\frac{x^2}{49} - \frac{9y^2}{343} = 1

.

Q15Exercise 10.4

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

Solution

Given: Foci are at

(0, \pm \sqrt{10})

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

To Find: The equation of the hyperbola.

Solution: Since the foci are on the y-axis, the transverse axis is along the y-axis. The standard equation is

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

. From the foci

(0, \pm c)

, we have

c = \sqrt{10}

. We know the relation

c^2 = a^2 + b^2

for a hyperbola.

(\sqrt{10})^2 = a^2 + b^2 \implies 10 = a^2 + b^2 \implies b^2 = 10 - a^2

Since the hyperbola passes through (2, 3), these coordinates must satisfy the equation:

\frac{3^2}{a^2} - \frac{2^2}{b^2} = 1

\frac{9}{a^2} - \frac{4}{b^2} = 1

Substitute

b^2 = 10 - a^2

into this equation:

\frac{9}{a^2} - \frac{4}{10 - a^2} = 1

Multiply 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

a^4 - 23a^2 + 90 = 0

Let

u = a^2

. The equation becomes

u^2 - 23u + 90 = 0

. Factorizing the quadratic equation:

(u - 5)(u - 18) = 0

So,

a^2 = 5

or

a^2 = 18

. Case 1: If

a^2 = 5

, then

b^2 = 10 - 5 = 5

. This is a valid solution. Case 2: If

a^2 = 18

, then

b^2 = 10 - 18 = -8

. Since

b^2

must be positive, this is not a valid solution. Therefore, we have

a^2 = 5

and

b^2 = 5

. Substituting these values into the standard equation:

\frac{y^2}{5} - \frac{x^2}{5} = 1

Final Answer: The equation of the hyperbola is

\frac{y^2}{5} - \frac{x^2}{5} = 1

or

y^2 - x^2 = 5

.

Q1Miscellaneous Exercise on Chapter 10

If a parabolic reflector is 20 cm in diameter and 5 cm deep, find the focus.

Solution

Given: A parabolic reflector with a diameter of 20 cm and a depth of 5 cm.

To Find: The focus of the reflector.

Solution: Let's orient the parabola with its vertex at the origin (0, 0) and its axis of symmetry along the positive x-axis. The equation of such a parabola is

y^2 = 4ax

, where the focus is at

(a, 0)

.

The reflector is 20 cm in diameter, which means the maximum value of

y

is

20/2 = 10

cm (and minimum is -10 cm). The depth of the reflector is 5 cm, which corresponds to the x-coordinate at the edge of the diameter. So, a point on the rim of the reflector has coordinates

(5, 10)

.

This point must satisfy the equation of the parabola

y^2 = 4ax

. Substitute

x = 5

and

y = 10

into the equation:

10^2 = 4a(5)

100 = 20a

a = \frac{100}{20} = 5

The focus of the parabola is at the point

(a, 0)

, which is

(5, 0)

. This means the focus is located 5 cm from the vertex along the axis of symmetry.

Final Answer: The focus is at a distance of 5 cm from the vertex of the reflector.

Q2Miscellaneous Exercise on Chapter 10

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

Solution

Given: A parabolic arch with a vertical axis. Height = 10 m. Width at the base = 5 m.

To Find: The width of the arch at a point 2 m from the vertex.

Solution: Let's set up a coordinate system with the vertex of the parabola at the origin (0, 0). Since the axis is vertical and the arch opens downwards, the equation of the parabola is of the form

x^2 = -4ay

.

The arch is 10 m high and 5 m wide at the base. This means the points at the base of the arch are located at a horizontal distance of

5/2 = 2.5

m from the center and a vertical distance of 10 m below the vertex. So, the points

(\pm 2.5, -10)

lie on the parabola.

Substitute the coordinates of one of these points, say

(2.5, -10)

, into the equation to find the value of

a

:

(2.5)^2 = -4a(-10)

6.25 = 40a

a = \frac{6.25}{40} = \frac{625}{4000} = \frac{1}{6.4} = \frac{5}{32}

So, the equation of the parabola is:

x^2 = -4\left(\frac{5}{32}\right)y = -\frac{5}{8}y

Now, we need to find the width of the arch at a height of 2 m from the vertex. This corresponds to a vertical distance of 2 m below the vertex, so

y = -2

. Substitute

y = -2

into the equation:

x^2 = -\frac{5}{8}(-2)

x^2 = \frac{10}{8} = \frac{5}{4}

x = \pm \sqrt{\frac{5}{4}} = \pm \frac{\sqrt{5}}{2}

The width of the arch at this height is the distance between the two points, which is

2|x|

. Width =

2 \times \frac{\sqrt{5}}{2} = \sqrt{5}

m.

Final Answer: The width of the arch 2 m from the vertex is

\sqrt{5}

meters.

Q3Miscellaneous Exercise on Chapter 10

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

Solution

Given: The bridge cable is a parabola. Roadway length = 100 m. Longest supporting wire = 30 m. Shortest supporting wire = 6 m.

To Find: The length of a supporting wire 18 m from the middle.

Solution: Let's set up a coordinate system with the vertex of the parabola at the lowest point of the cable. The equation of the parabola is of the form

x^2 = 4ay

.

The shortest wire is 6 m, which is at the middle of the bridge. This means the vertex of the parabola is 6 m above the roadway. The roadway is 100 m long, so it extends from

x = -50

m to

x = 50

m. The longest wires are at the ends of the roadway, at

x = \pm 50

m. Their length is 30 m.

The height of the cable at the ends (

x = \pm 50

) is 30 m above the roadway. Since the vertex is 6 m above the roadway, the y-coordinate of the ends of the cable relative to the vertex is

30 - 6 = 24

m. So, the points

(\pm 50, 24)

lie on the parabola.

Substitute the coordinates of the point (50, 24) into the equation

x^2 = 4ay

to find

a

:

50^2 = 4a(24)

2500 = 96a

a = \frac{2500}{96} = \frac{625}{24}

So, the equation of the parabola is:

x^2 = 4\left(\frac{625}{24}\right)y = \frac{625}{6}y

We need to find the length of the supporting wire 18 m from the middle, which is at

x = 18

m. Substitute

x = 18

into the equation to find the height

y

of the cable above the vertex:

18^2 = \frac{625}{6}y

324 = \frac{625}{6}y

y = \frac{324 \times 6}{625} = \frac{1944}{625} \approx 3.11\,\text{m}

This is the height of the cable above its lowest point. The total length of the supporting wire is this height plus the height of the vertex above the roadway (the shortest wire length). Length of wire =

y + 6 = \frac{1944}{625} + 6 = \frac{1944 + 3750}{625} = \frac{5694}{625} = 9.1104\,\text{m}

.

Final Answer: The length of the supporting wire attached to the roadway 18 m from the middle is 9.11 m (approximately).

Q4Miscellaneous Exercise on Chapter 10

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

Solution

Given: A semi-elliptical arch. Width = 8 m. Height at the centre = 2 m.

To Find: The height of the arch at a point 1.5 m from one end.

Solution: Let's place the centre of the ellipse at the origin (0, 0). The major axis is along the x-axis. The equation of the ellipse is

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

.

The width of the arch is 8 m, which corresponds to the length of the major axis. So,

2a = 8 \implies a = 4

. The height at the centre is 2 m, which corresponds to the length of the semi-minor axis. So,

b = 2

.

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 (

y

) at a point 1.5 m from one end. The ends of the arch are at

x = \pm 4

. A point 1.5 m from one end (say, from

x=4

) is at an x-coordinate of:

x = 4 - 1.5 = 2.5 \,\text{m}

Now, substitute

x = 2.5

into the equation of the ellipse to find the corresponding height

y

:

\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} = \frac{39}{16}

y = \sqrt{\frac{39}{16}} = \frac{\sqrt{39}}{4}

Since height must be positive, we take the positive root.

y \approx \frac{6.245}{4} \approx 1.56

m.

Final Answer: The height of the arch at a point 1.5 m from one end is

\frac{\sqrt{39}}{4}

meters.

Q5Miscellaneous Exercise on Chapter 10

A rod of length 12 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 cm from the end in contact with the $x$ -axis.

Solution

Given: A rod of length 12 cm with its ends on the coordinate axes. A point P on the rod, 3 cm from the end in contact with the x-axis.

To Find: The equation of the locus of point P.

Solution: Let the rod be AB, where end A is on the x-axis and end B is on the y-axis. Let the coordinates of A be

(p, 0)

and the coordinates of B be

(0, q)

. The length of the rod is 12 cm, so the distance between A and B is 12. Using the distance formula:

\sqrt{(p - 0)^2 + (0 - q)^2} = 12 \implies p^2 + q^2 = 12^2 = 144

.

Let the coordinates of point P be

(x, y)

. The point P is on the rod AB and is 3 cm from the end A (the end on the x-axis). This means that P divides the line segment AB in the ratio

AP:PB = 3:(12-3) = 3:9 = 1:3

.

Using the section formula for the coordinates of P:

x = \frac{3 \cdot p + 1 \cdot 0}{1 + 3} = \frac{3p}{4} \implies p = \frac{4x}{3}

y = \frac{3 \cdot 0 + 1 \cdot q}{1 + 3} = \frac{q}{4} \implies q = 4y

Now, substitute these expressions for

p

and

q

into the equation

p^2 + q^2 = 144

:

\left(\frac{4x}{3}\right)^2 + (4y)^2 = 144

\frac{16x^2}{9} + 16y^2 = 144

Divide the entire equation by 16:

\frac{x^2}{9} + y^2 = 9

To write it in the standard form of an ellipse, divide by 9:

\frac{x^2}{81} + \frac{y^2}{9} = 1

This is the equation of an ellipse.

Final Answer: The equation of the locus of point P is

\frac{x^2}{9} + y^2 = 9

or

\frac{x^2}{81} + \frac{y^2}{9} = 1

.

Q6Miscellaneous Exercise on Chapter 10

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

Solution

Given: A parabola with the equation

x^2 = 12y

.

To Find: The area of the triangle formed by the vertex and the endpoints of the latus rectum.

Solution: The equation of the parabola is

x^2 = 12y

. This is of the form

x^2 = 4ay

, which represents a parabola opening upwards with its vertex at the origin. Comparing the equations, we get:

4a = 12 \implies a = 3

.

Vertex: The vertex of the parabola is V(0, 0).
Focus: The focus is at F(0, a), so F is at (0, 3).
Latus Rectum: The latus rectum is a line segment passing through the focus and parallel to the directrix (which is $y = -3$ ). So, the equation of the line containing the latus rectum is $y = 3$ .
Endpoints of the Latus Rectum: To find the endpoints, substitute $y = 3$ into the parabola's equation: $x^2 = 12(3) = 36$ $x = \pm 6$ So, the endpoints of the latus rectum are L(-6, 3) and R(6, 3).

Now, we need to find the area of the triangle VLR with vertices V(0, 0), L(-6, 3), and R(6, 3). We can take the segment LR as the base of the triangle. Base length = Distance between L and R =

6 - (-6) = 12

units. The height of the triangle is the perpendicular distance from the vertex V(0, 0) to the base LR (which lies on the line

y=3

). Height = 3 units.

Area of

\triangle VLR = \frac{1}{2} \times \text{base} \times \text{height}

Area =

\frac{1}{2} \times 12 \times 3 = 18

square units.

Final Answer: The area of the triangle is 18 square units.

Q7Miscellaneous Exercise on Chapter 10

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

Solution

Given: The sum of the distances of a man from two fixed flag posts is always 10 m. The distance between the flag posts is 8 m.

To Find: The equation of the path traced by the man.

Solution: The locus of a point, the sum of whose distances from two fixed points is constant, is an ellipse. The two fixed points (flag posts) are the foci of the ellipse.

Let the two flag posts be the foci,

F_1

and

F_2

. The path of the man is an ellipse. Distance between the foci,

2c = 8

m, so

c = 4

m. The sum of the distances from any point on the ellipse to the foci is constant and equal to the length of the major axis,

2a

. So,

2a = 10

m, which means

a = 5

m.

Let's set up a coordinate system with the centre of the ellipse at the origin (0, 0) and the foci on the x-axis. The coordinates of the foci would be

F_1(-4, 0)

and

F_2(4, 0)

. The standard equation for such an ellipse is

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

.

We have

a = 5

and

c = 4

. We can find

b

using the relation

c^2 = a^2 - b^2

for an ellipse.

b^2 = a^2 - c^2

b^2 = 5^2 - 4^2 = 25 - 16 = 9

So,

a^2 = 25

and

b^2 = 9

.

Substituting these values into the standard equation:

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

Final Answer: The equation of the path traced by the man is

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

.

Q8Miscellaneous Exercise on Chapter 10

An equilateral triangle is inscribed in the parabola $y^{2}=4 a x$ , where one vertex is at the vertex of the parabola. Find the length of the side of the triangle.

Solution

Given: An equilateral triangle is inscribed in the parabola

y^2 = 4ax

. One vertex of the triangle is at the vertex of the parabola.

To Find: The length of the side of the triangle.

Solution: The vertex of the parabola

y^2 = 4ax

is at V(0, 0). Let this be one vertex of the equilateral triangle. Since the parabola is symmetric about the x-axis, for the triangle to be equilateral with one vertex at V, the other two vertices must be symmetric with respect to the x-axis. Let these vertices be P

(x_1, y_1)

and Q

(x_1, -y_1)

.

Let the side length of the equilateral triangle be

s

. Then, VP = VQ = PQ =

s

.

Distance PQ =

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

(assuming

y_1 > 0

). So,

s = 2y_1

.

Distance VP =

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

. So,

s^2 = x_1^2 + y_1^2

.

Since

s = 2y_1

, we have

s^2 = (2y_1)^2 = 4y_1^2

. Equating the two expressions for

s^2

:

x_1^2 + y_1^2 = 4y_1^2

x_1^2 = 3y_1^2 \quad ...(1)

The point P

(x_1, y_1)

lies on the parabola

y^2 = 4ax

. So, we have:

y_1^2 = 4ax_1 \quad ...(2)

Now, substitute equation (2) into equation (1):

x_1^2 = 3(4ax_1)

x_1^2 = 12ax_1

Since P is not the vertex,

x_1 \neq 0

. We can divide by

x_1

:

x_1 = 12a

Now find

y_1^2

using equation (2):

y_1^2 = 4a(12a) = 48a^2

y_1 = \sqrt{48a^2} = 4\sqrt{3}a

(taking the positive root for P).

The length of the side of the triangle is

s = 2y_1

.

s = 2(4\sqrt{3}a) = 8\sqrt{3}a

Final Answer: The length of the side of the equilateral triangle is

8\sqrt{3}a

.

Conic SectionsClass 11 Mathematics NCERT Solutions

In each of the following Exercises 1 to 5, find the equation of the circle with centre (0,2)(0,2)(0,2) and radius 2

Solution

centre (−2,3)(-2,3)(−2,3) and radius 4

Solution

centre (12,14)\left(\frac{1}{2}, \frac{1}{4}\right)(21​,41​) and radius 112\frac{1}{12}121​

Solution

centre (1,1)(1,1)(1,1) and radius 2\sqrt{2}2​

Solution

centre (−a,−b)(-a,-b)(−a,−b) and radius a2−b2\sqrt{a^{2}-b^{2}}a2−b2​

Solution

In each of the following Exercises 6 to 9, find the centre and radius of the circles. 6. (x+5)2+(y−3)2=36(x+5)^{2}+(y-3)^{2}=36(x+5)2+(y−3)2=36

Solution

x2+y2−4x−8y−45=0x^{2}+y^{2}-4 x-8 y-45=0x2+y2−4x−8y−45=0

Solution

x2+y2−8x+10y−12=0x^{2}+y^{2}-8 x+10 y-12=0x2+y2−8x+10y−12=0

Solution

2x2+2y2−x=02 x^{2}+2 y^{2}-x=02x2+2y2−x=0

Solution

Find the equation of the circle passing through the points (4,1)(4,1)(4,1) and (6,5)(6,5)(6,5) and whose centre is on the line 4x+y=164 x+y=164x+y=16.

Solution

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

Solution

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

Solution

Find the equation of the circle passing through (0,0)(0,0)(0,0) and making intercepts aaa and bbb on the coordinate axes.

Solution

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

Solution

Does the point (−2.5,3.5)(-2.5, 3.5)(−2.5,3.5) lie inside, outside or on the circle x2+y2=25x^{2}+y^{2}=25x2+y2=25 ?

Solution

In each of the following Exercises 1 to 6, find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum. y2=12xy^{2}=12 xy2=12x

Solution

x2=6yx^{2}=6 yx2=6y

Solution

y2=−8xy^{2}=-8 xy2=−8x

Solution

x2=−16yx^{2}=-16 yx2=−16y

Solution

y2=10xy^{2}=10 xy2=10x

Solution

x2=−9yx^{2}=-9 yx2=−9y

Solution

In each of the Exercises 7 to 12, find the equation of the parabola that satisfies the given conditions: 7. Focus (6,0)(6,0)(6,0); directrix x=−6x=-6x=−6

Solution

Focus (0,−3)(0,-3)(0,−3); directrix y=3y=3y=3

Solution

Vertex (0,0)(0,0)(0,0); focus (3,0)(3,0)(3,0)

Solution

Vertex (0,0)(0,0)(0,0); focus (−2,0)(-2,0)(−2,0)

Solution

Vertex (0,0)(0,0)(0,0) passing through (2,3)(2,3)(2,3) and axis is along xxx-axis.

Solution

Vertex (0,0)(0,0)(0,0), passing through (5,2)(5,2)(5,2) and symmetric with respect to yyy-axis.

Solution

In each of the Exercises 1 to 9, find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse. x236+y216=1\frac{x^{2}}{36}+\frac{y^{2}}{16}=136x2​+16y2​=1

Solution

x24+y225=1\frac{x^{2}}{4}+\frac{y^{2}}{25}=14x2​+25y2​=1

Solution

x216+y29=1\frac{x^{2}}{16}+\frac{y^{2}}{9}=116x2​+9y2​=1

Solution

x225+y2100=1\frac{x^{2}}{25}+\frac{y^{2}}{100}=125x2​+100y2​=1

Solution

x249+y236=1\frac{x^{2}}{49}+\frac{y^{2}}{36}=149x2​+36y2​=1

Solution

x2100+y2400=1\frac{x^{2}}{100}+\frac{y^{2}}{400}=1100x2​+400y2​=1

Solution

36x2+4y2=14436 x^{2}+4 y^{2}=14436x2+4y2=144

Solution

16x2+y2=1616 x^{2}+y^{2}=1616x2+y2=16

Solution

4x2+9y2=364 x^{2}+9 y^{2}=364x2+9y2=36

Solution

In each of the following Exercises 10 to 20, find the equation for the ellipse that satisfies the given conditions: 10. Vertices (±5,0)( \pm 5,0)(±5,0), foci (±4,0)( \pm 4,0)(±4,0)

Solution

Vertices (0,±13)(0, \pm 13)(0,±13), foci (0,±5)(0, \pm 5)(0,±5)

Solution

Vertices (±6,0)( \pm 6,0)(±6,0), foci (±4,0)( \pm 4,0)(±4,0)

Solution

Ends of major axis (±3,0)( \pm 3,0)(±3,0), ends of minor axis (0,±2)(0, \pm 2)(0,±2)

In each of the following Exercises 1 to 5, find the equation of the circle with centre $(0,2)$ and radius 2

centre $(-2,3)$ and radius 4

centre $\left(\frac{1}{2}, \frac{1}{4}\right)$ and radius $\frac{1}{12}$

centre $(1,1)$ and radius $\sqrt{2}$

centre $(-a,-b)$ and radius $\sqrt{a^{2}-b^{2}}$

In each of the following Exercises 6 to 9, find the centre and radius of the circles. 6. $(x+5)^{2}+(y-3)^{2}=36$

$x^{2}+y^{2}-4 x-8 y-45=0$

$x^{2}+y^{2}-8 x+10 y-12=0$

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

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

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

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

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

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

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

In each of the following Exercises 1 to 6, find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum. $y^{2}=12 x$

$x^{2}=6 y$

$y^{2}=-8 x$

$x^{2}=-16 y$

$y^{2}=10 x$

$x^{2}=-9 y$

In each of the Exercises 7 to 12, find the equation of the parabola that satisfies the given conditions: 7. Focus $(6,0)$ ; directrix $x=-6$

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

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

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

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

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

In each of the Exercises 1 to 9, find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse. $\frac{x^{2}}{36}+\frac{y^{2}}{16}=1$

$\frac{x^{2}}{4}+\frac{y^{2}}{25}=1$

$\frac{x^{2}}{16}+\frac{y^{2}}{9}=1$

$\frac{x^{2}}{25}+\frac{y^{2}}{100}=1$

$\frac{x^{2}}{49}+\frac{y^{2}}{36}=1$

$\frac{x^{2}}{100}+\frac{y^{2}}{400}=1$

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

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

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

In each of the following Exercises 10 to 20, find the equation for the ellipse that satisfies the given conditions: 10. Vertices $( \pm 5,0)$ , foci $( \pm 4,0)$

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

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

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

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

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

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

Foci $( \pm 3,0), a=4$

$b=3, c=4$ , centre at the origin; foci on the $x$ axis.

Centre at $(0,0)$ , major axis on the $y$ -axis and passes through the points $(3,2)$ and $(1,6)$ .

Major axis on the $x$ -axis and passes through the points $(4,3)$ and $(6,2)$ .

In each of the Exercises 1 to 6, find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbolas. $\frac{x^{2}}{16}-\frac{y^{2}}{9}=1$

$\frac{y^{2}}{9}-\frac{x^{2}}{27}=1$

$9 y^{2}-4 x^{2}=36$

$16 x^{2}-9 y^{2}=576$

$5 y^{2}-9 x^{2}=36$

$49 y^{2}-16 x^{2}=784$ .

In each of the Exercises 7 to 15, find the equations of the hyperbola satisfying the given conditions. 7. Vertices $( \pm 2,0)$ , foci $( \pm 3,0)$

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

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

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

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

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

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

vertices $( \pm 7,0), e=\frac{4}{3}$ .

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

A rod of length 12 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 cm from the end in contact with the $x$ -axis.

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

An equilateral triangle is inscribed in the parabola $y^{2}=4 a x$ , where one vertex is at the vertex of the parabola. Find the length of the side of the triangle.