Punjab State Board PSEB 11th Class Maths Book Solutions Chapter 11 Conic Sections Ex 11.2 Textbook Exercise Questions and Answers.

## PSEB Solutions for Class 11 Maths Chapter 11 Conic Sections Ex 11.2

Question 1.

Find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length .of the latus rectum for y^{2} = 12x.

Answer.

The given equation is y^{2} = 12x.

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

4a = 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 = 4a = 4 × 3 = 12.

Question 2.

Find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum for x^{2} = 6y.

Answer.

The given equation is x^{2} = 6y.

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

On comparing this equation with x^{2} = 4ay , we obtain

4a = 6

⇒ a = \(\frac{3}{2}\).

Coordinates of the focus = (0, a) = (0, \(\frac{3}{2}\))

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

Equation of directrix, y = – a i.e., y = – \(\frac{3}{2}\)

Length of latus rectum = 4a = 6.

Question 3.

Find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum for y^{2} = – 8x.

Answer.

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

– 4a = – 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 = 4a = 8.

Question 4.

Find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum for x^{2} = – 16y.

Ans.

The given equation is x^{2} = – 16y

Here, the coefficient of y is negative.

Hence, the parabola opens downwards.

On comparing this equation with x^{2} = – 4ay, we obtain

– 4a = – 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 cf directrix, y = a i.e., y = 4

Length of latus rectum = 4a = 16.

Question 5.

Find the coordinntes of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum for y^{2} = 10x.

Answer.

The given equation is y^{2} = 10x.

Here, the coefficient of x is positive.

Hence, the parabola opens towards the right.

On comparing this equation with y^{2} = 4ax, we obtain

4a = 10

⇒ a = \(\frac{5}{2}\)

Coordinates of the focus = (a, 0) = (\(\frac{5}{2}\), 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 = – \(\frac{5}{2}\)

Length of latus rectum = 4a = 10.

Question 6.

Find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum for x\(\frac{5}{2}\) = – 9y.

Answer.

We have, equation of parabola is x^{2} = – 9y

So, axis of symmetry is along x-axis.

The coefficient of y is negative, so parabola opens downward.

On comparing with the equation,

x^{2} = – 4ay, we get a = \(\frac{9}{4}\)

Thus, the focus is (0, – a) i.e., (0, – \(\frac{9}{4}\))

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

Length of latus rectum = 4a i.e., 9.

Question 7.

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

Answer.

Given, focus (6, 0) and directrix, x = – 6.

Here, we see that in focus, x-coordinate is positive and y-coordinate is zero.

So, focus lies on the positive direction of X-axis i.e., equation of parabola will be of the form y^{2} = 4ax with a = 6.

Hence, required equation is

y^{2} = 4 × 6x

⇒ y^{2} = 24x

Question 8.

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

Answer.

Given, 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} = – 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 x^{2}= – 4ay.

Here, a = 3.

Thus, the equation of the parabola is x^{2} = – 12y.

Question 9.

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

Answer.

Given, 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} = 4ax.

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

Thus, the equation of the parabola is y^{2} = 4 × 3 × x, i.e., y^{2} = 12x.

Question 10.

Find the equation of the parabola that satisfies the given conditions: Vertex (0, 0) focus (-2, 0).

Answer.

Given, 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} = – 4ax.

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

Thus, the equation of the parabola is y^{2} = – 4(2)x, i.e., y = – 8x.

Question 11.

Find the equation of the parabola that satisfies the given 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} = 4ax or y^{2} = – 4ax.

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} = 4ax.

3^{2} = 4a(2)

a = \(\frac{9}{8}\)

Thus, the equation of the paraola is y^{2} = – 4(2)x, i.e., y^{2} = – 8x.

Question 12.

Find the equation of the parabola that statisfies the given conditions; Vertex (0, 0), passing through (5, 2) and symmetric with respect to y-axis.

Answer.

Since, the parabola is symmetric about Y-axis and has its vertex at the origin, so the equation is of the form x^{2} = 4ay or x^{2} = – 4ay.

But parabola passes through (5, 2) which lies in the first quadrant. It must open upward.

Thus, the equation is of the form

x^{2} = 4ay ……………….(i)

Since, parabola passes through (5, 2)

(5)^{2} = 4a(2)

⇒ 25 = 8a

⇒ a = \(\frac{25}{8}\)

On putting the value of a in eq.(i), we get

x^{2} = 4 (\(\frac{25}{8}\)) y

⇒ x^{2} = \(\frac{25}{8}\) y