# NCERT solution class 11 chapter 11 Conic Sections exercise 11.2 mathematics

## EXERCISE 11.2

#### Question 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

The given equation is y2 = 12x.

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

On comparing this equation with y= 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 = –i.e., = – 3 i.e., x + 3 = 0

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

#### Question 2:

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

The given equation is x2 = 6y.

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

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

∴Coordinates of the focus = (0, a) =

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

Equation of directrix,

Length of latus rectum = 4a = 6

#### Question 3:

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

The given equation is y2 = –8x.

Here, the coefficient of 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

#### Question 4:

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

The given equation is x2 = –16y.

Here, the coefficient of 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

#### Question 5:

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

The given equation is y2 = 10x.

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

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

∴Coordinates of the focus = (a, 0)

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

Equation of directrix,

Length of latus rectum = 4a = 10

#### Question 6:

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

The given equation is x2 = –9y.

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

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

∴Coordinates of the focus =

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

Equation of directrix,

Length of latus rectum = 4a = 9

#### Question 7:

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

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.

#### Question 8:

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

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

x= – 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.

#### Question 9:

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

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

#### Question 10:

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

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

#### Question 11:

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

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.

Thus, the equation of the parabola is

#### Question 12:

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

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.

Thus, the equation of the parabola is

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