## NCERT Solutions for Class 11 Maths Chapter 11 – Conic Sections Ex 11.2

#### Page No 246:

#### Question 1:

Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of the latus rectum for

*y*^{2}= 12*x*#### Answer:

The given equation is

*y*^{2}= 12*x*.
Here, the coefficient of

*x*is positive. Hence, the parabola opens towards the right.
On comparing this equation with

*y*^{2 }= 4*ax*, we obtain
4

*a*= 12 ⇒*a*= 3
∴Coordinates of the focus = (

*a*, 0) = (3, 0)
Since the given equation involves

*y*^{2}, the axis of the parabola is the*x*-axis.
Equation of directrix,

*x*= –*a*i.e.,*x*= – 3 i.e.,*x*+ 3 = 0
Length of latus rectum = 4

*a*= 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

*x*^{2}= 6*y*#### Answer:

The given equation is

*x*^{2}= 6*y*.
Here, the coefficient of

*y*is positive. Hence, the parabola opens upwards.
On comparing this equation with

*x*^{2}= 4*ay*, we obtain
∴Coordinates of the focus = (0,

*a*) =
Since the given equation involves

*x*^{2}, the axis of the parabola is the*y*-axis.
Equation of directrix,

Length of latus rectum = 4

*a*= 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

*y*^{2}= – 8*x*#### Answer:

The given equation is

*y*^{2}= –8*x*.
Here, the coefficient of

*x*is negative. Hence, the parabola opens towards the left.
On comparing this equation with

*y*^{2}= –4*ax*, we obtain
–4

*a*= –8 ⇒*a*= 2
∴Coordinates of the focus = (–

*a*, 0) = (–2, 0)
Since the given equation involves

*y*^{2}, the axis of the parabola is the*x*-axis.
Equation of directrix,

*x*=*a*i.e.,*x*= 2
Length of latus rectum = 4

*a*= 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

*x*^{2}= – 16*y*#### Answer:

The given equation is

*x*^{2}= –16*y*.
Here, the coefficient of

*y*is negative. Hence, the parabola opens downwards.
On comparing this equation with

*x*^{2}*= –*4*ay,*we obtain
–4

*a*= –16 ⇒*a*= 4
∴Coordinates of the focus = (0, –

*a*) = (0, –4)
Since the given equation involves

*x*^{2}, the axis of the parabola is the*y*-axis.
Equation of directrix,

*y*=*a*i.e.,*y*= 4
Length of latus rectum = 4

*a*= 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

*y*^{2}= 10*x*#### Answer:

The given equation is

*y*^{2}= 10*x*.
Here, the coefficient of

*x*is positive. Hence, the parabola opens towards the right.
On comparing this equation with

*y*^{2 }= 4*ax*, we obtain
∴Coordinates of the focus = (

*a*, 0)
Since the given equation involves

*y*^{2}, the axis of the parabola is the*x*-axis.
Equation of directrix,

Length of latus rectum = 4

*a*= 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

*x*^{2}= –9*y*#### Answer:

The given equation is

*x*^{2}= –9*y*.
Here, the coefficient of

*y*is negative. Hence, the parabola opens downwards.
On comparing this equation with

*x*^{2}= –4*ay*, we obtain
∴Coordinates of the focus =

Since the given equation involves

*x*^{2}, the axis of the parabola is the*y*-axis.
Equation of directrix,

Length of latus rectum = 4

*a*= 9#### Page No 247:

#### Question 7:

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

*x*= –6#### Answer:

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

*y*^{2}= 4*ax*or*y*

^{2}= – 4

*ax*.

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*y*^{2}= 4*ax*.
Here,

*a*= 6
Thus, the equation of the parabola is

*y*^{2}= 24*x*.#### Question 8:

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

*y*= 3#### Answer:

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

*x*^{2}= 4*ay*or*x*

^{2 }= – 4

*ay*.

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*x*^{2}= –4*ay*.
Here,

*a*= 3
Thus, the equation of the parabola is

*x*^{2}= –12*y*.#### Question 9:

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

#### Answer:

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*y*^{2}= 4*ax*.
Since the focus is (3, 0),

*a*= 3.
Thus, the equation of the parabola is

*y*^{2}= 4 × 3 ×*x*, i.e.,*y*^{2}= 12*x*#### Question 10:

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

#### Answer:

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*y*^{2}= –4*ax*.
Since the focus is (–2, 0),

*a*= 2.
Thus, the equation of the parabola is

*y*^{2}= –4(2)*x*, i.e.,*y*^{2}= –8*x*#### 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#### Answer:

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*y*^{2}= 4*ax*or*y*^{2}= –4*ax*.
The parabola passes through point (2, 3), which lies in the first quadrant.

Therefore, the equation of the parabola is of the form

*y*^{2}= 4*ax*, while point
(2, 3) must satisfy the equation

*y*^{2}= 4*ax*.
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#### Answer:

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*x*^{2}= 4*ay*or*x*^{2}= –4*ay*.
The parabola passes through point (5, 2), which lies in the first quadrant.

Therefore, the equation of the parabola is of the form

*x*^{2}= 4*ay*, while point
(5, 2) must satisfy the equation

*x*^{2}= 4*ay*.
Thus, the equation of the parabola is

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