Class 11 Maths Ncert Solutions Chapter 11 Ex 11.2 Conic Sections PDF

# Class 11 Maths Ncert Solutions Ex 11.2

## Class 11 Maths Ncert Solutions Chapter 11 Ex 11.2

Q.1: For the equation y2 = 16x, determine the focus coordinates, the directrix equation, the length of the latus rectum and the axis of the parabola.

Sol:

Given:

y2 = 16x

As the coefficient of x is positive, the given parabola has the opening on the right side.

By comparing this equation with y2 = 4ax, we get:

4a = 16 ⇒ a = 4

Therefore, focus coordinates = (a, 0) = (4, 0)

The given equation has y2, the x-axis is the axis of the parabola.

Directrix equation, x = – a = – 4

So, x + 4 = 0

Therefore, Length of the latus rectum = 4a = 4 × 4 = 16

Q.2: For the equation x2 = 8y, determine the focus coordinates, the directrix equation, the length of the latus rectum and the axis of the parabola.

Sol:

Given:

x2 = 8y

As the coefficient of y is positive, the given parabola has the opening upwards.

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

4a = 8 ⇒ a = 2

Therefore, focus coordinates = (0, a) = (0, 2)

The given equation has x2, the y – axis is the axis of the parabola.

Directrix equation, y = – a = – 2

So, x + 2 = 0

Therefore, Length of the latus rectum = 4a = 4 × 2 = 8

Q.3: For the equation y2 = – 12x, determine the focus coordinates, the directrix equation, the length of the latus rectum and the axis of the parabola.

Sol:

Given:

y2 = – 12x

As the coefficient of x is negative, the given parabola has the opening on the left side.

By comparing this equation with y2 = – 4ax, we get

4a = 12 ⇒ a = 3

Therefore, focus coordinates = (- a, 0) = (- 3, 0)

The given equation has y2, the x-axis is the axis of the parabola.

Directrix equation, x = a = 2

So, x + 4 = 0

Therefore, Length of the latus rectum = 4a = 4 × 3 = 12

Q.4: For the equation x2 = – 20y, determine the focus coordinates,  the directrix equation, the length of the latus rectum and the axis of the parabola.

Sol:

Given:

x2 = – 20y

As the coefficient of y is negative, the given parabola has the opening downwards.

By comparing this equation with x2 = – 4ay, we get:

– 4a = – 20 ⇒ a = 5

Therefore, focus coordinates = (0, a) = (0, 5)

The given equation has x2, the y – axis is the axis of the parabola.

Directrix equation, y =  a = 5

Therefore, Length of the latus rectum = 4a = 4 × 5 = 20

Q.5: For the equation y2 = 24x, determine the focus coordinates, the directrix equation, the length of the latus rectum and the axis of the parabola.

Sol:

Given:

y2 = 24x

As the coefficient of x is positive, the given parabola has the opening on the right side.

By comparing this equation with y2 = 4ax, we get:

4a = 24 ⇒ a = 6

Therefore, focus coordinates = (a, 0) = (6, 0)

The given equation has y2, the x-axis is the axis of the parabola.

Directrix equation, x = – a = – 6

So, x + 6 = 0

Therefore, Length of the latus rectum = 4a = 4 × 6 = 24

Q.6: For the equation x2 = – 7y, determine the focus coordinates,  the directrix equation, the length of the latus rectum and the axis of the parabola.

Sol:

Given:

x2 = – 7y

As the coefficient of y is negative, the given parabola has the opening downwards.

By comparing this equation with x2 = – 4ay, we get:

– 4a = – 7 ⇒ a = 74$\frac{7}{4}$

Therefore, focus coordinates = (0, a) = (0, 74$\frac{7}{4}$)

The given equation has x2, the y – axis is the axis of the parabola.

Directrix equation, y = a = 74$\frac{7}{4}$

Therefore, Length of the latus rectum = 4a = 4 × 74$\frac{7}{4}$ = 7

Q7. Obtain the equation of the parabola which satisfies the given conditions below:

(i) Focus coordinates = (8, 0)

(ii) Directrix (x) = – 8

Sol:

Given:

(i) Focus coordinates = (8, 0)

(ii) Directrix (x) = – 8

The axis of the given parabola is the x – axis as the focus is represented on the x – axis.

The required equation can be either y2 = 4ax and y2 = – 4ax

As we know, directrix (x) = – 8

As the coefficient of x is negative, the given parabola has the opening on the left side and the focus (8, 0)

Therefore, the equation of the parabola = y2 = 4ax,

Here, a = 8

Thus, the equation of the parabola = y2 = 32 x.

Q.8: Obtain the equation of the parabola which satisfies the given conditions below:

(i) Focus coordinates = (0, – 5)

(ii) Directrix (y) = 5

Sol:

Given:

(i) Focus coordinates = (0, – 5)

(ii) Directrix (x) = 5

The axis of the given parabola is the y – axis as the focus is represented on the y – axis.

The required equation can be either x2 = 4ay and x2 = – 4ay

As we know, directrix (x) = 5

As the coefficient of x is negative, the given parabola has the opening on the left side and the focus (0, -5)

Therefore, the equation of the parabola = x2 = – 4ay,

Here, a = 5

Thus, the equation of the parabola = x2 = – 20y

Q.9: Obtain the equation of the parabola which satisfies the given conditions below:

(i) Vertex is at origin.

(ii) Focus coordinates = (4, 0)

Sol:

Given:

(i) Vertex is at origin i.e., (0, 0)

(ii) Focus coordinates = (4, 0)

As the vertex of a parabola lies at origin and the focus is represented on the positive side of the x – axis, and hence, the axis of the parabola is x – axis.

The required equation is y2 = 4ax

As we know, focus is (4, 0)

Therefore, the equation of the parabola = y2 = 4ax

Here, a = 4

Thus, the equation of the parabola = y2 = 16 x

Q.10: Obtain the equation of the parabola which satisfies the given conditions below:

(i) Vertex is at origin.

(ii) Focus coordinates = (- 3, 0)

Sol:

Given:

(i) Vertex is at origin i.e., (0, 0)

(ii) Focus coordinates = (- 3, 0)

As the vertex of a parabola lies at origin and the focus is represented on the negative side of the x – axis, and hence, the axis of the parabola is x – axis.

The required equation is y2 = – 4ax

Therefore, the equation of the parabola = y2 = – 4ax

As we know, focus is (- 3, 0)

Here, a = 3

Thus, the equation of the parabola = y2 = – 12 x

Q.11: Obtain the equation of the parabola which satisfies the given conditions below:

(i) Vertex is at origin.

(ii) Passing through coordinates = (3, 5)

The axis of the parabola is on x – axis.

Sol:

Given:

(i) Vertex is at origin i.e., (0, 0)

(ii) Passing through coordinates = (3, 5)

As the vertex of a parabola lies at origin and the passing through coordinates = (3, 5).

The required equation is either y2 = 4ax or y2 = – 4ax

The coordinates (3, 5) is in first quadrant.

Therefore, the equation of the parabola = y2 = 4ax, whereas (3, 5) should satisfy the equation

So,

52 = 4a (3)

25 = 12 a

a =  2512$\frac{25}{12}$

The parabola’s equation is:

y2 = 4 (2512$\frac{25}{12}$) x

3 y2 = 25 x

Q.12: Obtain the equation of the parabola which satisfies the given conditions below:

(i) Vertex is at origin.

(ii) Passing through coordinates = (6, 4)

The equation is symmetric as considered with y – axis.

Sol:

Given:

(i) Vertex is at origin i.e., (0, 0)

(ii) Passing through coordinates = (6, 4)

As the vertex of a parabola lies at origin and the passing through coordinates = (6, 4).

The required equation is either x2 = 4ay or x2 = – 4ay

The coordinates (6, 4) is in first quadrant.

Therefore, the equation of the parabola = x2 = 4ay, whereas (6, 4) should satisfy the equation

So,

62 = 4a (4)

36 = 16 a

a =  3616$\frac{36}{16}$

a =  94$\frac{9}{4}$

The parabola’s equation is:

x2 = 4(94$\frac{9}{4}$)y

x2 = 9 y