NCERT Solutions Class 11th Maths Chapter – 11 Conic Sections Exercise 11.2

NCERT Solutions Class 11th Maths Chapter – 11 Conic Sections Exercise 11.2

NCERT Solutions Class 11th Maths Chapter – 11 Conic Sections Exercise 11.2

?Chapter – 11?

✍Conic Sections✍

?Exercise 11.2?

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.

1. y² = 12x

Solution: 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.
∴ The equation of directrix, x = -a, then,
x + 3 = 0
Length of latus rectum = 4a = 4 × 3 = 12

2. x² = 6y
Solution: 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.
∴ The equation of directrix, y =-a, then,
y = -3/2
Length of latus rectum = 4a = 4(3/2) = 6

3. y² = – 8x

Solution: 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

4. x² = – 16y

Solution: 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.
∴ The equation of directrix, y =a, then,
y = 4
Length of latus rectum = 4a = 4(4) = 16

5. y2 = 10x

Solution: 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.
∴ The equation of directrix, x =-a, then,
x = – 5/2
Length of latus rectum = 4a = 4(5/2) = 10

6. x² = – 9y

Solution: 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.
∴ The equation of directrix, y = a, then,
y = 9/4
Length of latus rectum = 4a = 4(9/4) = 9
In each of the Exercises 7 to 12, find the equation of the parabola that satisfies the given conditions:

7. Focus (6,0); directrix x = – 6

Solution: Given:
Focus (6,0) and directrix x = -6
We know that the focus lies on the x–axis is the axis of the parabola.
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 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
∴ The equation of the parabola is y² = 24x.

8. Focus (0,–3); directrix y = 3

Solution: Given:
Focus (0, -3) and directrix y = 3
We know that the focus lies on the y–axis, the y-axis is the axis 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
∴ The equation of the parabola is x² = -12y.

9. Vertex (0, 0); focus (3, 0)

Solution: Given:
Vertex (0, 0) and focus (3, 0)
We know that 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.]
The equation of the parabola is of the form y² = 4ax.
Since, the focus is (3, 0), a = 3
∴ The equation of the parabola is y² = 4 × 3 × x,
y² = 12x

10. Vertex (0, 0); focus (–2, 0)

Solution: Given:
Vertex (0, 0) and focus (-2, 0)
We know that 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.]
The equation of the parabola is of the form y²=-4ax.
Since, the focus is (-2, 0), a = 2
∴ The equation of the parabola is y² = -4 × 2 × x,
y² = -8x

11. Vertex (0, 0) passing through (2, 3) and axis is along x-axis.

Solution: We know that the vertex is (0, 0) and the axis of the parabola is the x-axis
The equation of the parabola is either of the from y² = 4ax or y² = -4ax.
Given that the parabola passes through point (2, 3), which lies in the first quadrant.
So, the equation of the parabola is of the form y² = 4ax, while point (2, 3) must satisfy the equation y² = 4ax.
Then,
32 = 4a(2)
32 = 8a
9 = 8a
a = 9/8
Thus, the equation of the parabola is
y² = 4 (9/8)x
= 9x/2
2y² = 9x
∴ The equation of the parabola is 2y2 = 9x

12. Vertex (0, 0), passing through (5, 2) and symmetric with respect to y-axis.

Solution: We know that the vertex is (0, 0) and the parabola is symmetric about the y-axis.
The equation of the parabola is either of the from x² = 4ay or x² = -4ay.
Given that the parabola passes through point (5, 2), which lies in the first quadrant.
So, the equation of the parabola is of the form x² = 4ay, while point (5, 2) must satisfy the equation x² = 4ay.
Then,
52 = 4a(2)
25 = 8a
a = 25/8
Thus, the equation of the parabola is
x² = 4 (25/8)y
x² = 25y/2
2×2 = 25y
∴ The equation of the parabola is 2×2 = 25y