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

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 y2 = 12x.
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 = 4 ax, 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 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 x2 = 6y.
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 = $$\frac{3}{2}$$.
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 y2 = – 8x.
Here, the coefficient of x is negative.
Hence, the parabola opens towards the left.
On comparing this equation with y2 = – 4 ax , 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 the directrix and the length of the latus rectum for x2 = – 16y.
Ans.
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 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 y2 = 10x.
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 = $$\frac{5}{2}$$
Coordinates of the focus = (a, 0) = ($$\frac{5}{2}$$, 0)
Since the given equation involves y2, the axis of the parabola is the x-axis.
Equation of directrix, x = – a, i.e., x = – $$\frac{5}{2}$$
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.
We have, equation of parabola is x2 = – 9y
So, axis of symmetry is along x-axis.
The coefficient of y is negative, so parabola opens downward.
On comparing with the equation,
x2 = – 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.
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 y2 = 4ax with a = 6.
Hence, required equation is
y2 = 4 × 6x
⇒ y2 = 24x

Question 8.
Find the equation of the parabola that satisfies the given conditions : Focus (0, – 3); directrix y = 3.
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 x2 = 4 ay 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.

Question 9.
Find the equation of the parabola that satisfies the given conditions : Vertex (0, 0); focus (3, 0).
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 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 given conditions: Vertex (0, 0) focus (-2, 0).
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 y2 = – 4ax.
Since the focus is (- 2, 0), a = 2.
Thus, the equation of the parabola is y2 = – 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.
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 = 4 ax, while point (2, 3) must satisfy the equation y2 = 4ax.
32 = 4a(2)
a = $$\frac{9}{8}$$
Thus, the equation of the paraola is y2 = – 4(2)x, i.e., y2 = – 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.
⇒ a = $$\frac{25}{8}$$
x2 = 4 ($$\frac{25}{8}$$) y
⇒ x2 = $$\frac{25}{8}$$ y