# 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 y ^{2} = 12x**

**Solution**

The given equation is y^{2} = 12x.

Here, the coefficient of x is positive. Hence, the parabola opens towards the right.

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

4a = 12

⇒ a = 3

∴Coordinates of the focus = (a, 0) = (3, 0)

Since the given equation involves y^{2}, 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 x ^{2} = 6y**

**Solution**

The given equation is x^{2} = 6y.

Here, the coefficient of y is positive. Hence, the parabola opens upwards.

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

4a = 6

⇒ a = 3/2

∴ Coordinates of the focus = (0, a) = (0, 3/2)

Since the given equation involves x^{2} , 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 y ^{2} = – 8x**

**Solution**

The given equation is y^{2} = –8x.

Here, the coefficient of x is negative. Hence, the parabola opens towards the left.

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

–4a = –8

⇒ a = 2

∴ Coordinates 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 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 x ^{2} = – 16y**

**Solution**

The given equation is x^{2} = –16y.

Here, the coefficient of y is negative. Hence, the parabola opens downwards.

On comparing this equation with x^{2} = – 4ay, we obtain

–4a = –16

⇒ a = 4

∴ Coordinates of the focus = (0, –a) = (0, –4)

Since the given equation involves x^{2}, 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 y ^{2} = 10x**

**Solution**

The given equation is y^{2} = 10x.

Here, the coefficient of x is positive. Hence, the parabola opens towards the right.

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

4a = 10

⇒ a = 5/2

∴ Coordinates of the focus = (a, 0) = (5/2, 0)

Since the given equation involves y^{2} , 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 x ^{2} = –9y .**

**Solution**

The given equation is x^{2} = –9y.

Here, the coefficient of y is negative. Hence, the parabola opens downwards.

On comparing this equation with x^{2} = –4ay, we obtain

-4a = -9

⇒ b = 9/4

∴ Coordinates of the focus = (0, -a) = (0, -9/4)

Since the given equation involves x^{2} , 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 y^{2} = 4ax or

y^{2} = – 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 y^{2} = 4ax.

Here, a = 6

Thus, the equation of the parabola is y^{2} = 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 *x*^{2} = 4*ay* or*x*^{2 }= – 4*ay*.

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} = –4*ay*.

Here, *a* = 3

Thus, the equation of the parabola is *x*^{2} = –12*y*.

**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 *y*^{2} = 4*ax*.

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

Thus, the equation of the parabola is *y*^{2} = 4×3×*x*, i.e., *y*^{2} = 12*x*

**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 y^{2} = –4ax.

Since the focus is (–2, 0), a = 2.

Thus, the equation of the parabola is y^{2} = –4(2)x, i.e., y^{2} = –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 y^{2} = 4ax or y^{2} = –4ax.

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

Therefore, the equation of the parabola is of the form y^{2} = 4ax, while point

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

∴ 3^{2} = 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 x^{2} = 4ay or x^{2} = –4ay.

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

Therefore, the equation of the parabola is of the form x^{2} = 4ay, while point

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

∴ (5)^{2} = 4×a ×2 ⇒ 25 = 8a ⇒ a = 25/8

Thus, the equation of the parabola is