NCERT Solutions For Class 11 Maths chapter-11 Conic Sections Exercise 11.2

NCERT Solutions For Class 11 Maths chapter-11 Conic Sections Exercise 11.2

NCERT Solutions for Class-11 Maths Chapter-11 Conic Sections

NCERT Solutions For Class 11 Maths Chapter-11 Conic Sections Exercise 11.2 is Prepared by expert of Physics Wallah score more with Physics Wallah NCERT Class 11 maths solutions. You can download and share NCERT Solutions for Class 11 Maths.

NCERT Solutions for Class-11 Maths Exercise 11.2

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.

Question 1. Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of the latus rectum for y ² = 12x

Solution :
The given equation is y ² = 12x.

Here, the coefficient of x is positive. Hence, the parabola opens towards the right.

On comparing this equation with y ² = 4ax, we obtain

4a = 12 ⇒ a = 3

∴Coordinates of the focus = (a, 0) = (3, 0)

Since the given equation involves y ² , the axis of the parabola is the x-axis.

Equation of direcctrix, x = –a i.e., x = – 3 i.e., x + 3 = 0

Length of latus rectum = 4a = 4 × 3 = 12

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

Solution :
The given equation is x ² = 6y.

Here, the coefficient of y is positive. Hence, the parabola opens upwards.

On comparing this equation with x ² = 4ay, we obtain

∴Coordinates of the focus = (0, a) =

Since the given equation involves x ² , the axis of the parabola is the y-axis.

Equation of directrix,

Length of latus rectum = 4a = 6

Question 3. Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of the latus rectum for y ² = – 8x

Solution :
The given equation is y ² = –8x.

Here, the coefficient of x is negative. Hence, the parabola opens towards the left.

On comparing this equation with y ² = –4ax, we obtain

–4a = –8 ⇒ a = 2

∴Coordinates of the focus = (–a, 0) = (–2, 0)

Since the given equation involves y ² , the axis of the parabola is the x-axis.

Equation of directrix, x = a i.e., x = 2

Length of latus rectum = 4a = 8

Question 4. Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of the latus rectum for x ² = – 16y

Solution :

The given equation is x ² = –16 y .

Here, the coefficient of y is negative. Hence, the parabola opens downwards.

On comparing this equation with x ² = – 4 ay, we obtain

– 4 a = –16 ⇒ a = 4

∴ Coordinates of the focus = (0, – a ) = (0, –4)

Since the given equation involves x ² , the axis of the parabola is the y -axis.

Equation of directrix , y = a i.e., y = 4

Length of latus rectum = 4 a = 16

Question 5. Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of the latus rectum for y ² = 10x

Solution :

The given equation is y ² = 10 x .

Here, the coefficient of x is positive. Hence, the parabola opens towards the right.

On comparing this equation with y ² = 4 ax , we obtain

∴ Coordinates of the focus = ( a , 0)

Since the given equation involves y ² , the axis of the parabola is the x -axis.

Equation of directrix,

Length of latus rectum = 4 a = 10

Question 6. Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of the latus rectum for x ² = –9y

Solution :

The given equation is x ² = –9 y .

Here, the coefficient of y is negative. Hence, the parabola opens downwards.

On comparing this equation with x ² = –4 ay , we obtain

∴ Coordinates of the focus =

Since the given equation involves x ² , the axis of the parabola is the y -axis.

Equation of directrix,

Length of latus rectum = 4 a = 9

In each of the following Exercises 7 to 12, find the equation of the parabola that satisfies the given conditions:

Question 7. Find the equation of the parabola that satisfies the following conditions: Focus (6, 0); directrix x = –6

Solution :
Focus (6, 0); directrix, x = –6

Since the focus lies on the x-axis, the x-axis is the axis of the parabola.

Therefore, the equation of the parabola is either of the form y ² = 4ax or

y ² = – 4ax.

It is also seen that the directrix, x = –6 is to the left of the y-axis, while the focus (6, 0) is to the right of the y-axis. Hence, the parabola is of the form y ² = 4ax.

Here, a = 6

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

Question 8. Find the equation of the parabola that satisfies the following conditions: Focus (0, –3); directrix y = 3

Solution :
Focus = (0, –3); directrix y = 3

Since the focus lies on the y-axis, the y-axis is the axis of the parabola.

Therefore, the equation of the parabola is either of the form x ² = 4ay or x ² = – 4ay.

It is also seen that the directrix, y = 3 is above the x-axis, while the focus

(0, –3) is below the x-axis. Hence, the parabola is of the form x ² = –4ay.

Here, a = 3

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

Question 9. Find the equation of the parabola that satisfies the following conditions: Vertex (0, 0); focus (3, 0)

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

Since the vertex of the parabola is (0, 0) and the focus lies on the positive x-axis, x-axis is the axis of the parabola, while the equation of the parabola is of the form y ² = 4ax.

Since the focus is (3, 0), a = 3.

Thus, the equation of the parabola is y ² = 4 × 3 × x,

i.e., y ² = 12x

Question 10. Find the equation of the parabola that satisfies the following conditions: Vertex (0, 0) focus (–2, 0)

Solution :
Vertex (0, 0) focus (–2, 0)

Since the vertex of the parabola is (0, 0) and the focus lies on the negative x-axis, x-axis is the axis of the parabola, while the equation of the parabola is of the form y ² = –4ax.

Since the focus is (–2, 0), a = 2.

Thus, the equation of the parabola is y ² = –4(2)x,

i.e., y ² = –8x

Question 11. Find the equation of the parabola that satisfies the following conditions: Vertex (0, 0) passing through (2, 3) and axis is along x-axis

Solution :

Since the vertex is (0, 0) and the axis of the parabola is the x -axis, the equation of the parabola is either of the form y ² = 4 ax or y ² = –4 ax .

The parabola passes through point (2, 3), which lies in the first quadrant.

Therefore, t he equation of the parabola is of the form y ² = 4 ax , while point

(2, 3) must satisfy the equation y ² = 4 ax .

Thus, the e quation of the parabola is

Question 12. Find the equation of the parabola that satisfies the following conditions: Vertex (0, 0), passing through (5, 2) and symmetric with respect to y-axis

Solution :
Given:

Since the vertex is (0, 0) and the parabola is symmetric about the y -axis, the equation of the parabola is either of the form x ² = 4 ay or x ² = –4 ay .

T he parabola passes through point (5, 2), which lies in the first quadrant.

Therefore, t he equation of the parabola is of the form x ² = 4 ay , while point

(5, 2) must satisfy the equation x ² = 4 ay .

Thus, the equation of the parabola is