Chapter 11 Conic Sections Ex-11.1 |
Chapter 11 Conic Sections Ex-11.3 |
Chapter 11 Conic Sections Ex-11.4 |

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

**Answer
1** :

Given:

The equation is y^{2} =12x

Here we know that the coefficient of x is positive.

So, the parabola opens towards the right.

On comparing this equation with y^{2} =4ax, we get,

4a = 12

a = 3

Thus, 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.

∴ Theequation of directrix, x = -a, then,

x + 3 = 0

Length of latus rectum = 4a = 4 × 3 = 12

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

**Answer
2** :

Given:

The equation is x^{2} =6y

Here we know that the coefficient of y is positive.

So, the parabola opens upwards.

On comparing this equation with x^{2} =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 x^{2},the axis of the parabola is the y-axis.

∴ Theequation of directrix, y =-a, then,

y = -3/2

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

**y ^{2} = – 8x**

**Answer
3** :

Given:

The equation is y^{2} =-8x

Here we know that the coefficient of x is negative.

So, the parabola open towards the left.

On comparing this equation with y^{2} =-4ax, we get,

-4a = -8

a = -8/-4 = 2

Thus, 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.

∴Equation of directrix, x =a, then,

x = 2

Length of latus rectum = 4a = 4 (2) = 8

**x ^{2} = – 16y**

**Answer
4** :

Given:

The equation is x^{2} =-16y

Here we know that the coefficient of y is negative.

So, the parabola opens downwards.

On comparing this equation with x^{2} =-4ay, we get,

-4a = -16

a = -16/-4

= 4

Thus, 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.

∴ Theequation of directrix, y =a, then,

y = 4

Length of latus rectum = 4a = 4(4) = 16

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

**Answer
5** :

Given:

The equation is y^{2} =10x

Here we know that the coefficient of x is positive.

So, the parabola open towards the right.

On comparing this equation with y^{2} =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 y^{2},the axis of the parabola is the x-axis.

∴ Theequation of directrix, x =-a, then,

x = – 5/2

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

**x ^{2} = – 9y**

**Answer
6** :

Given:

The equation is x^{2} =-9y

Here we know that the coefficient of y is negative.

So, the parabola open downwards.

On comparing this equation with x^{2} =-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 x^{2},the axis of the parabola is the y-axis.

∴ Theequation of directrix, y = a, then,

y = 9/4

Length of latus rectum = 4a = 4(9/4) = 9

Focus (6,0); directrix x = – 6

**Answer
7** :

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

We know that the focus lies on the x–axis is the axis of theparabola.

So, the equation of the parabola is either of the form y^{2} = 4ax or y^{2} =-4ax.

It is also seen that the directrix, x = -6 is to the left of they- axis,

While the focus (6, 0) is to the right of the y –axis.

Hence, the parabola is of the form y^{2} =4ax.

Here, a = 6

∴ Theequation of the parabola is y^{2} =24x.

Focus (0,–3); directrix y = 3

**Answer
8** :

Given:

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

We know that the focus lies on the y–axis, the y-axis is theaxis of the parabola.

So, the equation of the parabola is either of the form x^{2} = 4ay or x^{2} =-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 x^{2} =-4ay.

Here, a = 3

∴ Theequation of the parabola is x^{2} =-12y.

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

**Answer
9** :

Given:

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

We know that the vertex of the parabola is (0, 0) and the focuslies on the positive x-axis. [x-axis is the axis of the parabola.]

The equation of the parabola is of the form y^{2 }= 4ax.

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

∴ Theequation of the parabola is y^{2} =4 × 3 × x,

y^{2} = 12x

Vertex (0, 0); focus (–2, 0)

**Answer
10** :

Given:

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

We know that the vertex of the parabola is (0, 0) and the focuslies on the positive x-axis. [x-axis is the axis of the parabola.]

The equation of the parabola is of the form y^{2}=-4ax.

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

∴ Theequation of the parabola is y^{2} =-4 × 2 × x,

y^{2} = -8x

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