NCERT Solutions for Class 11 Maths Chapter 11- Conic Sections Exercise 11.2

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Here, we discuss the second exercise of the chapter conic section. The Exercise 11.2 of NCERT Solutions for Class 11 Maths Chapter 11- Conic Sections is based on the following topics:

  1. Parabola
    1. Standard equations of parabola
    2. Latus rectum

Download PDF of NCERT Solutions for Class 11 Maths Chapter 11- Conic Sections Exercise 11.2

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Solutions for Class 11 Maths Chapter 11 – 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.


1. y2 = 12x

Solution:

Given:

The equation is y2 = 12x

Here we know that the coefficient of x is positive.

So, the parabola opens towards the right.

On comparing this equation with y2 = 4ax, we get,

4a = 12

a = 3

Thus, the co-ordinates of the focus = (a, 0) = (3, 0)

Since, the given equation involves y2, the axis of the parabola is the x-axis.

∴ The equation of directrix, x = -a, then,

x + 3 = 0

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

2. x2 = 6y

Solution:

Given:

The equation is y2 = 6y

Here we know that the coefficient of y is positive.

So, the parabola opens upwards.

On comparing this equation with y2 = 4ay, we get,

4a = 6

a = 6/4

= 3/2

Thus, the co-ordinates of the focus = (0,a) = (0, 3/2)

Since, the given equation involves x2, the axis of the parabola is the y-axis.

∴ The equation of directrix, y =-a, then,

y = -3/2

Length of latus rectum = 4a = 4(3/2) = 6

3. y2 = – 8x

Solution:

Given:

The equation is y2 = -8x

Here we know that the coefficient of x is negative.

So, the parabola open towards the left.

On comparing this equation with y2 = -4ax, we get,

-4a = -8

a = -8/-4 = 2

Thus, co-ordinates of the focus = (-a,0) = (-2, 0)

Since, the given equation involves y2, the axis of the parabola is the x-axis.

∴ Equation of directrix, x =a, then,

y = 2

Length of latus rectum = 4a = 4 (2) = 8

4. x2 = – 16y

Solution:

Given:

The equation is x2 = -16y

Here we know that the coefficient of y is negative.

So, the parabola opens downwards.

On comparing this equation with x2 = -4ay, we get,

-4a = -16

a = -16/-4

= 4

Thus, co-ordinates of the focus = (0,-a) = (0,-4)

Since, the given equation involves x2, the axis of the parabola is the y-axis.

∴ The equation of directrix, x =a, then,

y = 4

Length of latus rectum = 4a = 4(4) = 16

5. y2 = 10x

Solution:

Given:

The equation is y2 = 10x

Here we know that the coefficient of x is positive.

So, the parabola open towards the right.

On comparing this equation with y2 = -4ax, we get,

4a = 10

a = 10/4 = 5/2

Thus, co-ordinates of the focus = (a,0) = (5/2, 0)

Since, the given equation involves y2, the axis of the parabola is the x-axis.

∴ The equation of directrix, x =-a, then,

x = – 5/2

Length of latus rectum = 4a = 4(5/2) = 10

6. x2 = – 9y

Solution:

Given:

The equation is x2 = -9y

Here we know that the coefficient of y is negative.

So, the parabola open downwards.

On comparing this equation with x2 = -4ay, we get,

-4a = -9

a = -9/-4 = 9/4

Thus, co-ordinates of the focus = (0,-a) = (0, 9/4)

Since, the given equation involves x2, the axis of the parabola is the y-axis.

∴ The equation of directrix, y = a, then,

x = 9/4

Length of latus rectum = 4a = 4(9/4) = 9

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 (6,0) and directrix x = -6

We know that the focus lies on the x–axis is the axis of the parabola.

So, the equation of the parabola is either of the form y2 = 4ax or y2 = -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 y2 = 4ax.

Here, a = 6

∴ The equation of the parabola is y2 = 24x.

8. Focus (0,–3); directrix y = 3

Solution:

Given:

Focus (0, -3) and directrix y = 3

We know that the focus lies on the y–axis, the y-axis is the axis of the parabola.

So, the equation of the parabola is either of the form x2 = 4ay or x2 = -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 x2 = -4ay.

Here, a = 3

∴ The equation of the parabola is x2 = -12y.

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

Solution:

Given:

Vertex (0, 0) and focus (3, 0)

We know that 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.]

The equation of the parabola is of the form y2 = 4ax.

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

∴ The equation of the parabola is y2 = 4 × 3 × x,

y2 = 12x

10. Vertex (0, 0); focus (–2, 0)

Solution:

Given:

Vertex (0, 0) and focus (-2, 0)

We know that 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.]

The equation of the parabola is of the form y2=-4ax.

Since, the focus is (-2, 0), a = 2

∴ The equation of the parabola is y2 = -4 × 2 × x,

y2 = -8x

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

Solution:

We know that the vertex is (0, 0) and the axis of the parabola is the x-axis

The equation of the parabola is either of the from y2 = 4ax or y2 = -4ax.

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

So, the equation of the parabola is of the form y2 = 4ax, while point (2, 3) must satisfy the equation y2 = 4ax.

Then,

32 = 4a(2)

32 = 8a

9 = 8a

a = 9/8

Thus, the equation of the parabola is

y2 = 4 (9/8)x

= 9x/2

2y2 = 9x

∴ The equation of the parabola is 2y2 = 9x

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

Solution:

We know that the vertex is (0, 0) and the parabola is symmetric about the y-axis.

The equation of the parabola is either of the from x2 = 4ay or x2 = -4ay.

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

So, the equation of the parabola is of the form x2 = 4ay, while point (5, 2) must satisfy the equation x2 = 4ay.

Then,

52 = 4a(2)

25 = 8a

a = 25/8

Thus, the equation of the parabola is

x2 = 4 (25/8)y

x2 = 25y/2

2x2 = 25y

∴ The equation of the parabola is 2x2 = 25y


Access Other Exercise Solutions of Class 11 Maths Chapter 11- Conic Sections

Exercise 11.1 Solutions 15 Questions

Exercise 11.3 Solutions 20 Questions

Exercise 11.4 Solutions 15 Questions

Miscellaneous Exercise On Chapter 11 Solutions 8 Questions

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