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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:

- Parabola
- Standard equations of parabola
- 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 â€“ 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. 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}Â = 4ax, we get,

4a = 12

a = 3

Thus, the co-ordinates 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 y^{2}Â = 6y

Here we know that the coefficient of y is positive.

So, the parabola opens upwards.

On comparing this equation with y^{2}Â = 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 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 open towards the left.

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

-4a = -8

a = -8/-4 = 2

Thus, co-ordinates 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,

y = 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}Â = -4ay, we get,

-4a = -16

a = -16/-4

= 4

Thus, co-ordinates 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, x =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 open towards the right.

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

4a = 10

a =Â 10/4 = 5/2

Thus, co-ordinates 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 open downwards.

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

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

x =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}Â = 4ax.

Here, a = 6

âˆ´ The equation of the parabola is y^{2}Â = 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 yâ€“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}Â = -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}Â = -4ay.

Here, a = 3

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

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

âˆ´ The equation of the parabola is y^{2}Â = 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.

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

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

y^{2}Â = -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 y^{2 }= 4ax or y^{2}Â = -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 y^{2}Â = 4ax, while point (2, 3) must satisfy the equation y^{2}Â = 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 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 x^{2 }= 4ay or x^{2}Â = -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 x^{2}Â = 4ay, while point (5, 2) must satisfy the equation x^{2}Â = 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- 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