Class 11 NCERT Solutions- Chapter 11 Conic Section – 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.

Question 1: y² = 12x

Solution:

Given equation: y² = 12x

Since the coefficient of x is positive.

Therefore, the parabola will open towards the right.

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

4a = 12

a = 3

Therefore, 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.

Thus, the equation of directrix, x = -a, therefore 

x = -3 

The length of latus rectum = 4a = 4 × 3 = 12

Question 2: x² = 6y

Solution:

Given equation : x² = 6y

Since the coefficient of y is positive.

Therefore, the parabola will open upwards.

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

4a = 6

a = 6/4

= 3/2

Therefore, 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.

Thus, the equation of directrix, y =-a, therefore,

y = -3/2

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

Question 3: y² = – 8x

Solution:

Given equation: y² = -8x

Since the coefficient of x is negative.

Therefore, the parabola will open towards the left.

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

-4a = -8

a = -8/-4 = 2

Therefore, the co-ordinates of the focus = (-a,0) = (-2, 0)

Since, the given equation involves y², 

The axis of the parabola is the x-axis.

Thus, the equation of directrix, x = a, therefore,

x = 2

The length of latus rectum = 4a = 4 (2) = 8

Question 4: x² = – 16y

Solution:

Given equation: x² = -16y

Since the coefficient of y is negative.

Therefore, the parabola will open downwards.

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

-4a = -16

a = -16/-4

= 4

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

Since, the given equation involves x², 

The axis of the parabola is the y-axis.

Thus, the equation of directrix, y =a, then,

y = 4

The length of latus rectum = 4a = 4(4) = 16

Question 5: y² = 10x

Solution:

Given equation: y² = 10x

Since the coefficient of x is positive.

Therefore, the parabola will open towards the right.

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

4a = 10

a = 10/4 = 5/2

Therefore, 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.

Thus, the equation of directrix, x = -a, then,

x = – 5/2

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

Question 6: x² = – 9y

Solution:

Given equation: x² = -9y

Since the coefficient of y is negative.

Therefore, the parabola will open downwards.

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

-4a = -9

a = -9/-4 = 9/4

Therefore, 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.

Thus, the equation of directrix, y = a, then,

y = 9/4

The 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:

Question 7: Focus (6,0); directrix x = – 6

Solution:

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

Since the focus lies on the x–axis 

The x-axis is the axis of the parabola.

As the directrix, x = -6 thus it is to the left of the y- axis,

The equation of the parabola is y² = 4ax

Here, a = 6

Therefore, the equation of the parabola is y² = 24x.

Question 8: Focus (0, –3); directrix y = 3

Solution:

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

Since the focus lies on the y–axis, 

The y-axis is the axis of the parabola.

As the given directrix, y = 3 thus it is above the x- axis,

The equation of the parabola is x² = -4ay

Here, a = 3

Therefore, the equation of the parabola is x² = -12y.

Question 9: Vertex (0, 0); focus (3, 0)

Solution:

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

Since the vertex of the parabola is (0, 0) 

The focus lies on the positive x-axis. 

The parabola is of the form y² = 4ax.

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

Therefore, the equation of the parabola is y² = 4 × 3 × x,

y² = 12x

Question 10: Vertex (0, 0); focus (–2, 0)

Solution:

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

Since the vertex of the parabola is (0, 0) 

The focus lies on the positive x-axis. 

The parabola is of the form y²=-4ax.

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

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

y² = -8x

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

Solution:

Given: Vertex is (0, 0) and the axis is along the x-axis

Also given that the parabola passes through the point (2, 3), which lies in the first quadrant.

Since equation of the parabola is y² = 4ax while the point (2, 3) must satisfy the equation y² = 4ax.

Therefore,

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

Therefore, the equation of the parabola is 2y² = 9x

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

Solution:

Given: Vertex is (0, 0) and the symmetric is with respect to the y-axis.

Also given that the parabola passes through the point (5, 2), which lies in the first quadrant.

Since, the equation of the parabola is x² = 4ay while the point (5, 2) must satisfy the equation x² = 4ay.

Therefore,

5² = 4a(2)

25 = 8a

a = 25/8

Thus, the equation of the parabola is

x² = 4 (25/8)y

x² = 25y/2

2x² = 25y

Therefore, the equation of the parabola is 2x² = 25y

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