# 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)