Conic Sections – Exercise 11.2 – Class XI

  1. Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of the latus rectum for y2 = 12x

Solution:

The given equation is y2 = 12x.

Here, the coefficient of x is positive. Hence, the parabola opens towards the right.

On comparing this equation with y2 = 4ax, we obtain

4a = 12

⇒ a = 3

∴ Coordinates of the focus = (a, 0) = (3, 0)

Since the given equation involves y2, the axis of the parabola is the x-axis.

Equation of direcctrix, x = –a i.e., x = – 3 i.e., x + 3 = 0

Length of latus rectum = 4a = 4 × 3 = 12


  1. Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of the latus rectum for x2 = 6y

Solution:

The given equation is x2 = 6y.

Here, the coefficient of y is positive. Hence, the parabola opens upwards.

On comparing this equation with x2 = 4ay, we obtain

4a = 6

a = 3/2

Coordinates of the focus = (0, a) = (0, 3/2)

Since the given equation involves x2, the axis of the parabola is the y-axis.

Equation of directrix, y = -a i.e., y = –3/2

Length of latus rectum = 4a = 6


  1. Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of the latus rectum for y2 = – 8x

Solution:

The given equation is y2 = –8x.

Here, the coefficient of x is negative. Hence, the parabola opens towards the left.

On comparing this equation with y2 = –4ax, we obtain

–4a = –8

⇒ a = 2

∴Coordinates of the focus = (–a, 0) = (–2, 0)

Since the given equation involves y2, the axis of the parabola is the x-axis.

Equation of directrix, x = a i.e., x = 2

Length of latus rectum = 4a = 8


  1. Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of the latus rectum for x2 = – 16y

Solution:

The given equation is x2 = –16y.

Here, the coefficient of y is negative. Hence, the parabola opens downwards.

On comparing this equation with x2 = – 4ay, we obtain

–4a = –16 ⇒ a = 4

∴Coordinates of the focus = (0, –a) = (0, –4)

Since the given equation involves x2, the axis of the parabola is the y-axis.

Equation of directrix, y = a i.e., y = 4

Length of latus rectum = 4a = 16


  1. Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of the latus rectum for y2 = 10x

Solution:

The given equation is y2 = 10x.

Here, the coefficient of x is positive. Hence, the parabola opens towards the right.

On comparing this equation with y2 = 4ax, we obtain

4a = 10

a = 5/2

Coordinates of the focus = (a, 0)

Since the given equation involves y2, the axis of the parabola is the x-axis.

Equation of directrix, x = -a i.e., x = –5/2

Length of latus rectum = 4a = 10


  1. Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of the latus rectum for x2 = –9y

Solution:

The given equation is x2 = –9y.

Here, the coefficient of y is negative. Hence, the parabola opens downwards.

On comparing this equation with x2 = –4ay, we obtain

Coordinates of the focus =(0, -a) = (0, –9/4)

Since the given equation involves x2, the axis of the parabola is the y-axis.

Equation of directrix, y =a i.e., y = 9/4

Length of latus rectum = 4a = 9


  1. Find the equation of the parabola that satisfies the following conditions: Focus (6, 0); directrix x = –6

Solution:

Focus (6, 0); directrix, x = –6

Since the focus lies on the x-axis, the x-axis is the axis of the parabola.

Therefore, the equation of the parabola is either of the form

y2 = 4ax

or

y2 = – 4ax.

It is also seen that the directrix, x = –6 is to the left of the y-axis, while the focus (6, 0) is to the right of the y-axis.

Hence, the parabola is of the form y2 = 4ax.

Here, a = 6

Thus, the equation of the parabola is y2 = 24x.


  1. Find the equation of the parabola that satisfies the following conditions: Focus (0, –3); directrix y = 3

Solution:

Focus = (0, –3); directrix y = 3

Since the focus lies on the y-axis, the y-axis is the axis of the parabola.

Therefore, the equation of the parabola is either of the form

x2 = 4ay

or

x2 = – 4ay.

It is also seen that the directrix, y = 3 is above the x-axis, while the focus (0, –3) is below the x-axis. Hence, the parabola is of the form x2 = –4ay.

Here, a = 3

Thus, the equation of the parabola is x2 = –12y.


  1. Find the equation of the parabola that satisfies the following conditions: Vertex (0, 0); focus (3, 0)

Solution:

Vertex (0, 0); focus (3, 0)

Since the vertex of the parabola is (0, 0) and the focus lies on the positive x-axis, x-axis is the axis of the parabola, while the equation of the parabola is of the form y2 = 4ax.

Since the focus is (3, 0), a = 3.

Thus, the equation of the parabola is y2 = 4 × 3 × x, i.e., y2 = 12x

 

  1. Find the equation of the parabola that satisfies the following conditions: Vertex (0, 0) focus (–2, 0)

Solution:

Vertex (0, 0) focus (–2, 0)

Since the vertex of the parabola is (0, 0) and the focus lies on the negative x-axis, x-axis is the axis of the parabola, while the equation of the parabola is of the form y2 = – 4ax.

Since the focus is (–2, 0), a = 2.

Thus, the equation of the parabola is y2 = –4(2)x, i.e., y2 = –8x


  1. Find the equation of the parabola that satisfies the following conditions: Vertex (0, 0) passing through (2, 3) and axis is along x-axis

Solution:

Since the vertex is (0, 0) and the axis of the parabola is the x-axis, the equation of the parabola is either of the form

y2 = 4ax

or

y2 = –4ax.

The parabola passes through point (2, 3), which lies in the first quadrant.

Therefore, the equation of the parabola is of the form y2 = 4ax, while point (2, 3) must satisfy the equation y2 = 4ax.

32 = 4a(2)

a = 9/8

Thus, the equation of the parabola is

y2 = 4(9/8)x

y2 = (9/2)x

2y2 = 9x


  1. Find the equation of the parabola that satisfies the following conditions: Vertex (0, 0), passing through (5, 2) and symmetric with respect to y-axis

Solution:

Since the vertex is (0, 0) and the parabola is symmetric about the y-axis, the equation of the parabola is either of the form

x2 = 4ay

or

x2 = –4ay.

The parabola passes through point (5, 2), which lies in the first quadrant.

Therefore, the equation of the parabola is of the form x2 = 4ay, while point (5, 2) must satisfy the equation x2 = 4ay.

52 = 4xax2

25 = 8a

a = 25/8

Thus, the equation of the parabola is

x2 = 4(25/8)y

2x2 = 25y