Chapter 11 – Conic Sections (Ex – 11.2)

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)