# 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: y**^{2} = 12x

^{2}= 12x

**Solution:**

Given equation: y

^{2}= 12xSince the coefficient of x is positive.

Therefore, the parabola will open towards the right.

While comparing this equation with y

^{2}= 4ax, we get,4a = 12

a = 3

Therefore, the co-ordinates of the focus = (a, 0) = (3, 0)

Since, the given equation involves y

^{2},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: x**^{2} = 6y

^{2}= 6y

**Solution:**

Given equation : x

^{2}= 6ySince the coefficient of y is positive.

Therefore, the parabola will open upwards.

While comparing this equation with x

^{2}= 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 x

^{2},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: y**^{2} = – 8x

^{2}= – 8x

**Solution:**

Given equation: y

^{2}= -8xSince the coefficient of x is negative.

Therefore, the parabola will open towards the left.

While comparing this equation with y

^{2}= -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 y

^{2},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: x**^{2} = – 16y

^{2}= – 16y

**Solution:**

Given equation: x

^{2}= -16ySince the coefficient of y is negative.

Therefore, the parabola will open downwards.

While comparing this equation with x

^{2}= -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 x

^{2},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: y**^{2} = 10x

^{2}= 10x

**Solution:**

Given equation: y

^{2}= 10xSince the coefficient of x is positive.

Therefore, the parabola will open towards the right.

While comparing this equation with y

^{2}= 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 y

^{2},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: x**^{2} = – 9y

^{2}= – 9y

**Solution:**

Given equation: x

^{2}= -9ySince the coefficient of y is negative.

Therefore, the parabola will open downwards.

While comparing this equation with x

^{2}= -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 x

^{2},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 y

^{2}= 4axHere, a = 6

Therefore, the equation of the parabola is y

^{2}= 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 x

^{2}= -4ayHere, a = 3

Therefore, the equation of the parabola is x

^{2}= -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 y

^{2}= 4ax.Since, the focus is (3, 0), a = 3

Therefore, the equation of the parabola is y

^{2}= 4 × 3 × x,y

^{2}= 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 y

^{2}=-4ax.Since, the focus is (-2, 0), a = 2

Therefore, the equation of the parabola is y

^{2}= -4 × 2 × x,y

^{2}= -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 y

^{2}= 4ax while the point (2, 3) must satisfy the equation y^{2}= 4ax.Therefore,

3

^{2}= 4a(2)3

^{2}= 8a9 = 8a

a = 9/8

Thus, the equation of the parabola is

y

^{2}= 4 (9/8)x= 9x/2

2y

^{2}= 9xTherefore, the equation of the parabola is 2y

^{2}= 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 x

^{2}= 4ay while the point (5, 2) must satisfy the equation x^{2}= 4ay.Therefore,

5

^{2}= 4a(2)25 = 8a

a = 25/8

Thus, the equation of the parabola is

x

^{2}= 4 (25/8)yx

^{2}= 25y/22x

^{2}= 25yTherefore, the equation of the parabola is 2x

^{2}= 25y

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