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

Students would be able to boost their confidence by solving the **NCERT Solutions for Class 11 Maths**. All the solutions are prepared by experts to help students excel in their studies. They can download these NCERT Solutions of Class 11 Maths and practise them to score well in the examinations.

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 â€“ Conic Sections is based on the following topics:

- Parabola
- Standard equations of parabola
- Latus rectum

## NCERT Solutions for Class 11 Maths Chapter 11 â€“ Conic Sections Exercise 11.2

### Solutions for Class 11 Maths Chapter 11 â€“ 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. y^{2}Â = 12x

**Solution:**

Given:

The equation is y^{2}Â = 12x

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

So, the parabola opens towards the right.

On comparing this equation with y^{2} = 4ax, we get

4a = 12

a = 3

Thus, the coordinates of the focus = (a, 0) = (3, 0)

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

âˆ´ The equation of directrix, x = -a, then

x + 3 = 0

Length of latus rectum = 4a = 4 Ã— 3 = 12

**2. x ^{2}Â = 6y**

**Solution:**

Given:

The equation is x^{2}Â = 6y

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

So, the parabola opens upwards.

On comparing this equation with x^{2} = 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 x^{2}, the axis of the parabola is the y-axis.

âˆ´ The equation of directrix, y =-a, then,

y = -3/2

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

**3. y ^{2}Â = â€“ 8x**

**Solution:**

Given:

The equation is y^{2}Â = -8x

Here, we know that the coefficient of x is negative.

So, the parabola opens towards the left.

On comparing this equation with y^{2} = -4ax, we get

-4a = -8

a = -8/-4 = 2

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

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

âˆ´ Equation of directrix, x =a, then

x = 2

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

**4. x ^{2}Â = â€“ 16y**

**Solution:**

Given:

The equation is x^{2}Â = -16y

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

So, the parabola opens downwards.

On comparing this equation with x^{2} = -4ay, we get

-4a = -16

a = -16/-4

= 4

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

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

âˆ´ The equation of directrix, y =a, then

y = 4

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

**5. y ^{2}Â = 10x**

**Solution:**

Given:

The equation is y^{2}Â = 10x

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

So, the parabola opens towards the right.

On comparing this equation with y^{2} = 4ax, we get

4a = 10

a =Â 10/4 = 5/2

Thus, coordinates of the focus = (a,0) = (5/2, 0)

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

âˆ´ The equation of directrix, x =-a, then

x = â€“ 5/2

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

**6. x ^{2}Â = â€“ 9y**

**Solution:**

Given:

The equation is x^{2}Â = -9y

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

So, the parabola opens downwards.

On comparing this equation with x^{2} = -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 x^{2}, the axis of the parabola is the y-axis.

âˆ´ The equation of directrix, y = a, then

y =Â 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 = â€“ 6

**Solution:**

Given:

Focus (6,0) and directrix x = -6

We know that the focus lies on the xâ€“axis is the axis of the parabola.

So, the equation of the parabola is either of the form y^{2}Â = 4ax or y^{2}Â = -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 â€“axis.

Hence, the parabola is of the form y^{2} = 4ax

Here, a = 6

âˆ´ The equation of the parabola is y^{2} = 24x

**8. Focus (0,â€“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 x^{2}Â = 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 form x^{2} = -4ay

Here, a = 3

âˆ´ The equation of the parabola is x^{2} = -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 y^{2 }= 4ax.

The focus is (3, 0), a = 3

âˆ´ The equation of the parabola is y^{2} = 4 Ã— 3 Ã— x

y^{2}Â = 12x

**10. Vertex (0, 0); focus (â€“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 y^{2}=-4ax

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

âˆ´ The equation of the parabola is y^{2} = -4 Ã— 2 Ã— x

y^{2}Â = -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 y^{2 }= 4ax or y^{2} = -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 y^{2} = 4ax, while points (2, 3) must satisfy the equation y^{2} = 4ax

Then,

3^{2}Â = 4a(2)

3^{2} = 8a

9 = 8a

a =Â 9/8

Thus, the equation of the parabola is

y^{2} = 4 (9/8)x

= 9x/2

2y^{2} = 9x

âˆ´ The equation of the parabola is 2y^{2} = 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 x^{2 }= 4ay or x^{2} = -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 x^{2} = 4ay, while points (5, 2) must satisfy the equation x^{2} = 4ay

Then,

5^{2}Â = 4a(2)

25 = 8a

a =Â 25/8

Thus, the equation of the parabola is

x^{2} = 4 (25/8)y

x^{2} = 25y/2

2x^{2} = 25y

âˆ´ The equation of the parabola is 2x^{2} = 25y

### Access Other Exercise Solutions of Class 11 Maths Chapter 11 â€“ 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 â€“ NCERT Class 11 Solutions

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