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Chapter 11 Conic Sections Ex-11.2 Interview Questions Answers

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Question 1 : 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.

y2 = 12x

Given:

The equation is y2 =12x

Here we know that the coefficient of x is positive.

So, the parabola opens towards the right.

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

4a = 12

a = 3

Thus, the co-ordinates of the focus = (a, 0) = (3, 0)

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

Theequation of directrix, x = -a, then,

x + 3 = 0

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

Question 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.

x2 = 6y

Given:

The equation is x2 =6y

Here we know that the coefficient of y is positive.

So, the parabola opens upwards.

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

4a = 6

a = 6/4

= 3/2

Thus, the co-ordinates of the focus = (0,a) = (0, 3/2)

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

Theequation of directrix, y =-a, then,

y = -3/2

Length of latus rectum = 4a = 4(3/2) = 6

Question 3 : 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.

y2 = – 8x

Given:

The equation is y2 =-8x

Here we know that the coefficient of x is negative.

So, the parabola open towards the left.

On comparing this equation with y2 =-4ax, we get,

-4a = -8

a = -8/-4 = 2

Thus, co-ordinates 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, then,

x = 2

Length of latus rectum = 4a = 4 (2) = 8

Question 4 : 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.

x2 = – 16y

Given:

The equation is x2 =-16y

Here we know that the coefficient of y is negative.

So, the parabola opens downwards.

On comparing this equation with x2 =-4ay, we get,

-4a = -16

a = -16/-4

= 4

Thus, co-ordinates of the focus = (0,-a) = (0,-4)

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

Theequation of directrix, y =a, then,

y = 4

Length of latus rectum = 4a = 4(4) = 16

Question 5 : 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.

y2 = 10x

Given:

The equation is y2 =10x

Here we know that the coefficient of x is positive.

So, the parabola open towards the right.

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

4a = 10

a = 10/4 = 5/2

Thus, co-ordinates of the focus = (a,0) = (5/2, 0)

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

Theequation of directrix, x =-a, then,

x = – 5/2

Length of latus rectum = 4a = 4(5/2) = 10

Question 6 : 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.

x2 = – 9y

Given:

The equation is x2 =-9y

Here we know that the coefficient of y is negative.

So, the parabola open downwards.

On comparing this equation with x2 =-4ay, we get,

-4a = -9

a = -9/-4 = 9/4

Thus, co-ordinates of the focus = (0,-a) = (0, -9/4)

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

Theequation of directrix, y = a, then,

y = 9/4

Length of latus rectum = 4a = 4(9/4) = 9

Question 7 : In each of the Exercises 7 to 12, find the equation of the parabola that satisfies the given conditions:
Focus (6,0); directrix x = – 6

Focus (6,0) and directrix x = -6

We know that the focus lies on the x–axis is the axis of theparabola.

So, 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 they- 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

Theequation of the parabola is y2 =24x.

Question 8 : In each of the Exercises 7 to 12, find the equation of the parabola that satisfies the given conditions:
Focus (0,–3); directrix y = 3

Given:

Focus (0, -3) and directrix y = 3

We know that the focus lies on the y–axis, the y-axis is theaxis of the parabola.

So, 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

Theequation of the parabola is x2 =-12y.

Question 9 : In each of the Exercises 7 to 12, find the equation of the parabola that satisfies the given conditions:
Vertex (0, 0); focus (3, 0)

Given:

Vertex (0, 0) and focus (3, 0)

We know that the vertex of the parabola is (0, 0) and the focuslies on the positive x-axis. [x-axis is the axis of the parabola.]

The equation of the parabola is of the form y= 4ax.

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

Theequation of the parabola is y2 =4 × 3 × x,

y2 = 12x

Question 10 : In each of the Exercises 7 to 12, find the equation of the parabola that satisfies the given conditions:
Vertex (0, 0); focus (–2, 0)

Given:

Vertex (0, 0) and focus (-2, 0)

We know that the vertex of the parabola is (0, 0) and the focuslies on the positive x-axis. [x-axis is the axis of the parabola.]

The equation of the parabola is of the form y2=-4ax.

Since, the focus is (-2, 0), a = 2

Theequation of the parabola is y2 =-4 × 2 × x,

y2 = -8x

krishan