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

*According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 10.

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Here, we discuss the second exercise of the chapter conic section. Chapter 11 Conic Sections of Class 11 Maths is categorised under the CBSE Syllabus for the academic session 2023-24. 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

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, the 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 coordinates 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 x² = 6y

Here, we know that the coefficient of y is positive.

So, the parabola opens upwards.

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

4a = 6

a = 6/4

= 3/2

Thus, the coordinates 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 opens towards the left.

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

-4a = -8

a = -8/-4 = 2

Thus, coordinates 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

x = 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, coordinates 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, y =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 opens towards the right.

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

4a = 10

a = 10/4 = 5/2

Thus, coordinates 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 opens downwards.

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

-4a = -9

a = -9/-4 = 9/4

Thus, coordinates 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

y = 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 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.

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

The focus is (-2, 0), a = 2

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

y² = -8x

11. Vertex (0, 0) passes through (2, 3), and the axis is along the 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 from y² = 4ax or y² = -4ax

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

So, the equation of the parabola is of the form y² = 4ax, while points (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 the 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 from x² = 4ay or x² = -4ay

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

So, the equation of the parabola is of form x² = 4ay, while points (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

Also explore – NCERT Class 11 Solutions