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

Chapter 11 Conic Sections Exercise 11.2 NCERT Solutions for Class 11 Maths will prove useful guide in completing your homework and improving your problem solving skills. Class 11 Maths NCERT Solutions are prepared by Studyrankers subject matter experts who have given in depth concepts in every question so students can easily understand them.

1. Find the coordinates of the focus, axis of the parabola, the equation of the directrix and length of latus rectum
y2 = 12x.

y 2 = 12x is the equation of parabola => 4a = 12
a = 3

Focus F (3, 0)
Axis is x-axis, i.e., y = 0
Directrix is x = –3
Length of Latus Rectum = 4a = 12

2. Find the coordinates of the focus, axis of the parabola, the equation of the directrix and length of latus rectum
x 2 = 6y.

x 2 = 6y is the equation of parabola
4a = 6,

∴ a = 6/4 = 3/2

Focus (0, 3/2)
Axis is y-axis i.e., x = 0
Directrix y = -(3/2)

3. Find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum.
y 2 = –8x.

The given equation involves y 2, so the axis of symmetry is along the x-axis.
The coefficient of x is negative, so the parabola opens to the left. Comparing with the given equation y 2 = 4ax, we find that a = 2. Thus the focus of the parabola is (–2, 0) and the equation of the directrix of the parabola is x = 2. Length of the Latus rectum LL' is 4a = 4 × 2 = 8

4. Find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum.
x 2 = –16y

The given equation involves x 2, so the axis of symmetry is along the y-axis.
The coefficient of y is negative, so the parabola opens downward. Comparing with the given equation x 2 = – 4ay, we find that a = 4. Thus the focus of the parabola is (0, –4) and the equation of the directrix of the parabola is y = 4. Length of the Latus rectum LL' is 4a = 4 × 4 = 16.

5. Find the coordinates of the focus, axis of the parabola, the equation of the directrix and length of latus rectum.
y 2 = 10x

y 2 = 10x is the equation of parabola
=> 4a = 10, a = 5/2

Focus (5/2 , 0)
Axis is x-axis, i.e., y = 0
Directrix is x = -(5/2)
Length of latus rectum = 10

6. Find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum for given curve
x 2 = –9y

Equation of the parabola is x 2 = –9y, 4a = 9, a = 9/4

Focus, (0, -(9/4))
Axis of parabola is Y-axis i.e x = 0, Directrix is y = 9/4
and length of latus rectum = 9.

7. Find the equation of the parabola that satisfies the given conditions:
Focus (6, 0); directrix x = –6.

Since the focus (6, 0) lies on the x-axis, the x-axis itself is the axis of the parabola. Hence the equation of the parabola is of the form of either y 2 = 4ax or y 2 = – 4ax. Since the directrix is x = – 6 and the focus is (6, 0) the parabola is to be of the form of y 2 = 4ax with a = 6. Hence, the required equation is y 2 = 4(6)x = 24x.

8. Find the equation of parabola that satisfies the given focus (0, –3) and directrix y = 3.

Focus (0, –3) and directrix y = 3

vertex is the mid point of (0, –3), (0, 3)
i.e. vertex is (0, 0)
and a = 3, ∴ 4a = 12
∴ Equation of parabola x 2 = –12y

9. Find the equation of the parabola that satisfies the given conditions:
Vertex (0, 0), Focus (3, 0)

Vertex (0, 0), Focus is (3, 0)
a = 3
∴ 4a = 12
∴ Equation of parabola is y 2 = 12x.

10. Find the equation of the parabola that satisfies the given conditions:
Vertex (0, 0), Focus (–2, 0).

Vertex (0, 0), Focus is (–2, 0)
a = 2, focus being (–2, 0) it is directed towards OX'.
∴ Equation of parabola is y 2 = –8x.

11. Find the equation of the parabola that satisfies the given conditions:
Vertex (0, 0), passing through (2, 3) and axis is along x-axis.

Since the parabola is symmetric about x-axis and has its vertex at the origin, the equation is of the form y 2 = 4ax or y 2 = – 4ax, where the sign depends on whether the parabola opens to the right or left. But the parabola passes through (2, 3) which lies in the first quadrant, it must open to the right. Thus the equation is of the form y 2 = 4ax.
Since the parabola passes through (2, 3) we have
32 = 4a(2) => 9 = 8a => a = 9/8
Therefore, the equation of the parabola is
y2 = 4(9/8)x = 9/2 x => 2y2 = 9x.

12. Find the equation of the parabola whose vertex (0, 0), passing through (5, 2) and symmetric with respect to y-axis.