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

NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Ex 11.2 are part of NCERT Solutions for Class 11 Maths. Here we have given NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Ex 11.2.

• Conic Sections Class 11 Ex 11.1
• Conic Sections Class 11 Ex 11.3
• Conic Sections Class 11 Ex 11.4

Board	CBSE
Textbook	NCERT
Class	Class 11
Subject	Maths
Chapter	Chapter 11
Chapter Name	Conic Sections
Exercise	Ex 11.2
Number of Questions Solved	12
Category	NCERT Solutions

NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Ex 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.

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.

Question 2.
x² = 6y
Solution:
The given equation of parabola is x² = 6y which is of the form x² = 4ay.

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.

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.

Question 5.
y²= 10x
Solution:
The given equation of parabola is y² = 10x, which is of the form y² = 4ax.

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:

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.

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.

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.

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.

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)

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)

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