# GSEB Solutions Class 11 Maths Chapter 11 Conic Sections Ex 11.2

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. y2 = 12x
2. x2 = 6y
3. y2 = – 8x
4. x2 = – 16y
5. y2 = 10x
6. x2 = – 9y
Solutions to questions 1-6:
1. y2 = 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. x2 = 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. y2 = – 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. x2 = – 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. y2 = 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
x2 = – 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 y2 = 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 x2 = – 12y.

9. Vertex (0, 0), focus is (3, 0).
So, a = 3 â‡’ 4a = 12.
âˆ´ Equation of parabola is y2 = 12x.

10. Vertex (0, 0), focus (- 2, 0).
a = 2, focus being (- 2, 0). It is directed towards OX’.
âˆ´ Equation of parabola is y2 = – 8x.

11. Vertex (0, 0). Parabola passes through (5, 2)
Axis of parabola is x-axis.
Let the equation of parabola be y2 = 4ax.
It passes through (2, 3).
âˆ´ 9 = 4.a.2
âˆ´ a = $$\frac{9}{8}$$.
âˆ´ Equation of parabola is y2 = $$\frac{9}{2}$$x or 2y2 = 9x.

12. Vertex (0, 0). Parabola passes through (5, 2).
It is symmetric about y-axis.
Let the equation of parabola be x2 = 4ay.
(5, 2) lies on it.
âˆ´ 25 = 4a.2. â‡’ a = $$\frac{25}{8}$$.
âˆ´ Equation of parabola is x2 = $$\frac{25}{2}$$y or 2x2 = 25y.