# NCERT Solutions For Class 11 Maths Chapter 11 Conic Sections Ex 11.2

Here, Below you all know about NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Ex 11.2 Question Answer. I know many of you confuse about finding Chapter 11 Conic Sections Ex 11.2 Of Class 11 NCERT Solutions. So, Read the full post below and get your solutions.

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

NCERT TEXTBOOK EXERCISES

Ex 11.2 Class 11 Maths Question 1.
y2= 12x

Solution:

The given equation of parabola is y2 = 12x which is of the form y2 = 4ax.
∴ 4a = 12 ⇒ a = 3
∴ Coordinates of focus are (3, 0)
Axis of parabola is y = 0
Equation of the directrix is x = -3 ⇒ x + 3 = 0
Length of latus rectum = 4 x 3 = 12.

Ex 11.2 Class 11 Maths Question 2.
x2 = 6y

Solution:

The given equation of parabola is x2 = 6y which is of the form x2 = 4ay.

Ex 11.2 Class 11 Maths Question 3.
y2 = – 8x

Solution:

The given equation of parabola is
y2 = -8x, which is of the form y2 = – 4ax.
∴ 4a = 8 ⇒ a = 2
∴ Coordinates of focus are (-2, 0)
Axis of parabola is y = 0
Equation of the directrix is x = 2 ⇒ x – 2 = 0
Length of latus rectum = 4 x 2 = 8.

Ex 11.2 Class 11 Maths Question 4.
x2 = -16y

Solution:

The given equation of parabola is
x2 = -16y, which is of the form x2 = -4ay.
∴ 4a = 16 ⇒ a = 4
∴ Coordinates of focus are (0, -4)
Axis of parabola is x = 0
Equation of the directrix is y = 4 ⇒ y – 4 = 0
Length of latus rectum = 4 x 4 = 16.

Ex 11.2 Class 11 Maths Question 5.
y2= 10x

Solution:

The given equation of parabola is y2 = 10x, which is of the form y2 = 4ax.

Ex 11.2 Class 11 Maths Question 6.
x2 = -9y

Solution:

The given equation of parabola is
x2 = -9y, which is of the form x2 = -4ay.

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

Ex 11.2 Class 11 Maths Question 7.
Focus (6, 0); directrix x = -6
Solution:

We are given that the focus (6, 0) lies on the x-axis, therefore x-axis is the axis of parabola. Also, the directrix is x = -6 i.e. x = -a and focus (6, 0) i.e. (a, 0). The equation of parabola is of the form y2 = 4ax.
The required equation of parabola is
y2 = 4 x 6x ⇒ y2 = 24x.

Ex 11.2 Class 11 Maths Question 8.
Focus (0, -3); directri xy=3

Solution:

We are given that the focus (0, -3) lies on the y-axis, therefore y-axis is the axis of parabola. Also the directrix is y = 3 i.e. y = a and focus (0, -3) i.e. (0, -a). The equation of parabola is of the form x2 = -4ay.
The required equation of parabola is
x2 = – 4 x 3y ⇒ x2 = -12y.

Ex 11.2 Class 11 Maths Question 9.
Vertex (0, 0); focus (3, 0)

Solution:

Since the vertex of the parabola is at (0, 0) and focus is at (3, 0)
∴ y = 0 ⇒ The axis of parabola is along x-axis
∴ The equation of the parabola is of the form y2 = 4ax
The required equation of the parabola is
y2 = 4 x 3x ⇒ y2 = 12x.

Ex 11.2 Class 11 Maths Question 10.
Vertex (0, 0); focus (-2, 0)

Solution:

Since the vertex of the parabola is at (0, 0) and focus is at (-2, 0).
∴ y = 0 ⇒ The axis of parabola is along x-axis
∴ The equation of the parabola is of the form y2 = – 4ax
The required equation of the parabola is
y2 = – 4 x 2x ⇒ y2 = -8x.

Ex 11.2 Class 11 Maths Question 11.
Vertex (0, 0), passing through (2, 3) and axis is along x-axis.

Solution:

Since the vertex of the parabola is at (0, 0) and the axis is along x-axis.
∴ The equation of the parabola is of the form y2 = 4ax
Since the parabola passes through point (2, 3)

Ex 11.2 Class 11 Maths Question 12.
Vertex (0, 0), passing through (5, 2) and symmetric with respect toy-axis.

Solution:

Since the vertex of the parabola is at (0, 0) and it is symmetrical about the y-axis.
∴ The equation of the parabola is of the form x2 = 4ay
Since the parabola passes through point (5, 2)

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

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