Class 11 Maths NCERT Solutions for Chapter 11 Conic Sections Exercise 11.2
Conic Sections Exercise 11.2 Solutions
1. Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of the latus rectum for y2 = 12x
Solution
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 = 4ax, 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 direcctrix, x = –a i.e., x = – 3 i.e., x + 3 = 0
Length of latus rectum = 4a = 4 × 3 = 12
2. Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of the latus rectum for x2 = 6y
Solution
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 = 3/2
∴ Coordinates of the focus = (0, a) = (0, 3/2)
Since the given equation involves x2 , the axis of the parabola is the y - axis.
Equation of directrix , y = -a i.e., y = -3/2
Length of latus rectum = 4a = 6
3. Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of the latus rectum for y2 = – 8x
Solution
The given equation is y2 = –8x.
Here, the coefficient of x is negative. Hence, the parabola opens towards the left.
On comparing this equation with y2 = –4ax, 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
4. Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of the latus rectum for x2 = – 16y
Solution
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 of directrix, y = a i.e., y = 4
Length of latus rectum = 4a = 16
5. Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of the latus rectum for y2 = 10x
Solution
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 = 5/2
∴ Coordinates of the focus = (a, 0) = (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 = -5/2
Length of latus rectum = 4a = 10
6. Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of the latus rectum for x2 = –9y .
Solution
The given equation is x2 = –9y.
Here, the coefficient of y is negative. Hence, the parabola opens downwards.
On comparing this equation with x2 = –4ay, we obtain
-4a = -9
⇒ b = 9/4
∴ Coordinates of the focus = (0, -a) = (0, -9/4)
Since the given equation involves x2 , the axis of the parabola is the y - axis.
Equation of directrix, y = a, i.e., y = 9/4
Length of latus rectum = 4a = 9
7. Find the equation of the parabola that satisfies the following conditions: Focus (6, 0); directrix x = –6
Solution
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 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
Thus, the equation of the parabola is y2 = 24x.
8. Find the equation of the parabola that satisfies the following conditions: Focus (0, –3); directrix y = 3
Solution
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 = 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
Thus, the equation of the parabola is x2 = –12y.
9. Find the equation of the parabola that satisfies the following conditions: Vertex (0, 0); focus (3, 0)
Solution
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
10. Find the equation of the parabola that satisfies the following conditions: Vertex (0, 0) focus (–2, 0)
Solution
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., y2 = –8x
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
Solution
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 = 4ax, while point
(2, 3) must satisfy the equation y2 = 4ax.
∴ 32 = 4a(2) ⇒ a = 9/8
Thus, the equation of the parabola is 
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
Solution
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 x2 = 4ay or x2 = –4ay.
The parabola passes through point (5, 2), which lies in the first quadrant.
Therefore, the equation of the parabola is of the form x2 = 4ay, while point
(5, 2) must satisfy the equation x2 = 4ay.
∴ (5)2 = 4×a ×2 ⇒ 25 = 8a ⇒ a = 25/8
Thus, the equation of the parabola is 