NCERT Solutions for Class 11 Maths Chapter 10 Conic Sections Ex 10.2

# NCERT Solutions for Class 11 Maths Chapter 10 Conic Sections Ex 10.2

NCERT Solutions for Class 11 Maths Chapter 10 Conic Sections Ex 10.2 are the part of NCERT Solutions for Class 11 Maths. Here you can find the NCERT Solutions for Class 11 Maths Chapter 10 Conic Sections Ex 10.2.

## NCERT Solutions for Class 11 Maths Chapter 10 Conic Sections Ex 10.2

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.

### Ex 10.2 Class 11 Maths Question 1.

y= 12x

Solution:
The given equation of parabola is:

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

### Ex 10.2 Class 11 Maths Question 2.

x2 = 6y

Solution:
The given equation of parabola is:

x2 = 6y which is of the form x2 = 4ay.

### Ex 10.2 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 the focus are (-2, 0).
Axis of parabola is y = 0.
Equation of the directrix is x = 2
x – 2 = 0
Length of the latus rectum = 4 × 2 = 8.

### Ex 10.2 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 the focus are (0, -4).
Axis of parabola is x = 0.
Equation of the directrix is y = 4
y – 4 = 0
Length of the latus rectum = 4 × 4 = 16.

### Ex 10.2 Class 11 Maths Question 5.

y= 10x

Solution:
The given equation of parabola is:

y2 = 10x, which is of the form y2 = 4ax.

### Ex 10.2 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 10.2 Class 11 Maths Question 7.

Focus (6, 0); directrix x = -6

Solution:
It is given that the focus (6, 0) lies on the x-axis, therefore, the 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.
Therefore, the required equation of the parabola is:
y2 = 4 × 6x
y2 = 24x

### Ex 10.2 Class 11 Maths Question 8.

Focus (0, -3); directrix y = 3

Solution:
It is 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.
Therefore, the required equation of the parabola is:
x2 = – 4 × 3y
x2 = -12y

### Ex 10.2 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 × 3x
y2 = 12x.

### Ex 10.2 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.
Therefore, the required equation of the parabola is:
y2 = –4 × 2x
y2 = -8x

### Ex 10.2 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 the point (2, 3).

### Ex 10.2 Class 11 Maths Question 12.

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