Gujarat Board GSEB Textbook Solutions Class 11 Maths Chapter 11 Conic Sections Ex 11.2 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 11 Maths Chapter 11 Conic Sections Ex 11.2

In each of the following questions 1 to 6, find the co-ordinates of the focus, axis of the parabola, the equation of the directrix and length of latus rectum:

Solutions to questions 1-6:

1. y^{2} = 12x

2. x^{2} = 6y

3. y^{2} = – 8x

4. x^{2} = – 16y

5. y^{2} = 10x

6. x^{2} = – 9y

Solutions to questions 1-6:

1. y^{2} = 12x is the equation of parabola.

So, 4a = 12 â‡’ a = 3.

âˆ´ Focus (3, 0)

Axis is x-axis, i.e., y = 0.

Directrix is x = – 3.

Length of latus rectum = 4a = 12.

2. x^{2} = 6y is the equation of parabola.

âˆ´ 4a = 6.

âˆ´ a = \(\frac{6}{4}\) = \(\frac{3}{2}\).

Focus (0, \(\frac{3}{2}\)).

Axis is y-axis i.e., x = 0.

Directrix is y = – \(\frac{3}{2}\).

Length of latus rectum = 6.

3. y^{2} = – 8x is the equation of parabola.

âˆ´ 4a = 8, â‡’ a = 2.

Focus (- 2, 0).

Axis is x-axis y = 0.

Latus rectum Directrix is x = 2.

Length of latus rectum = 4a = 8.

4. x^{2} = – 16y is the equation of parabola.

âˆ´ 4a = 16 â‡’ a = 4.

Focus (0, – 4).

Axis is y-axis, x = 0.

Directrix is y = 4.

Length of latus rectum = 16.

5. y^{2} = 10x is the equation of parabola.

âˆ´ 4a = 10 â‡’ a = \(\frac{5}{2}\).

Focus (\(\frac{5}{2}\), 0).

Axis is x-axis, i.e., y = 0.

Directrix is x = – \(\frac{5}{2}\).

Length of latus rectum = 10.

6. Equation of parabola is

x^{2} = – 9y, (see fig. 4)

âˆ´ 4a = 9 â‡’ a = \(\frac{9}{4}\).

Focus (0, – \(\frac{9}{4}\)).

Axis of parabola is y-axis i.e., x = 0.

Directrix is y = \(\frac{9}{4}\).

Length of latus rectum = 9.

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

7. Focus is (6, 0) and directrix is x = – 6.

8. Focus is (0, – 3), dlrectrix y = 3.

9. Vertex (0, 0) and Focus (3, 0).

10. Vertex (0, 0) and Focus (- 2, 0).

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

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

Solutions to questions 7-12:

7. Focus is (6, 0) and directrix is x = – 6. Its vertex is the mid-point of (6, 0) and (- 6, 0).

âˆ´ Vertex = (0, 0)

âˆ´ a = 6 and so equation of parabola is y^{2} = 24x.

8. Focus (0, – 3), directrix is y = 3.

Vertex is the mid-point of (0, – 3) and (0, 3).

i.e., vertex is (0, 0)

and a = 3 â‡’ 4a = 12.

âˆ´ Equation of parabola is x^{2} = – 12y.

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

So, a = 3 â‡’ 4a = 12.

âˆ´ Equation of parabola is y^{2} = 12x.

10. Vertex (0, 0), focus (- 2, 0).

a = 2, focus being (- 2, 0). It is directed towards OX’.

âˆ´ Equation of parabola is y^{2} = – 8x.

11. Vertex (0, 0). Parabola passes through (5, 2)

Axis of parabola is x-axis.

Let the equation of parabola be y^{2} = 4ax.

It passes through (2, 3).

âˆ´ 9 = 4.a.2

âˆ´ a = \(\frac{9}{8}\).

âˆ´ Equation of parabola is y^{2} = \(\frac{9}{2}\)x or 2y^{2} = 9x.

12. Vertex (0, 0). Parabola passes through (5, 2).

It is symmetric about y-axis.

Let the equation of parabola be x^{2} = 4ay.

(5, 2) lies on it.

âˆ´ 25 = 4a.2. â‡’ a = \(\frac{25}{8}\).

âˆ´ Equation of parabola is x^{2} = \(\frac{25}{2}\)y or 2x^{2} = 25y.