NCERT Solutions for Exercise 11.2 Class 11 Maths Chapter 11

NCERT Solutions for Exercise 11.2 Class 11 Maths Chapter 11 - Conic Section

Edited By Ravindra Pindel | Updated on Jul 18, 2022 11:48 AM IST

In the previous exercise, you have already learned about the section of the cone called ‘circle’. In the NCERT solutions for Class 11 Maths chapter 11 exercise 11.2, you will learn about another conic section called a parabola. The word parabola is made of two words ‘para’ which means ‘for’ and ‘bola’ which means ‘throwing’. The shape of the parabola is described by the path of a ball throwing in the air in the presence of gravity. The parabola is a set of points in a plane that are equidistant from a fixed line and a fixed point (not on the line) in the plane.

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Conic Section Class 11 Chapter 11 Exercise 11.2
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Benefits of NCERT Solutions for Class 11 Maths Chapter 11 Exercise 11.2:-
NCERT Solutions of Class 11 Subject Wise
NCERT Solutions for Class 11 Maths
Subject Wise NCERT Exampler Solutions

Also, you will learn about the standard equation of parabola, latus rectum of parabola, the focus of the parabola, directrix and vertex of a parabola, etc in the Class 11th Maths chapter 11 exercise 11.2. Parabola has a wide range of applications in the field of science, physics, research, design of ballistic missiles, etc. Class 11 Maths chapter 11 exercise 11.2 is very important for the CBSE exam as well as engineering entrance exams like JEE Main, SRMJEE, etc. If you are looking for NCERT solutions from Class 6 to 12 for Science and Maths at one place, click on the NCERT Solutions link.

Also, see

Conic Section Class 11 Chapter 11 Exercise 11.2

Question:1 Find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum.

$y^2 =12x$

Answer:

Given, a parabola with equation

$y^2 =12x$

This is parabola of the form $y^2=4ax$ which opens towards the right.

So,

By comparing the given parabola equation with the standard equation, we get,

$4a=12$

$a=3$

Hence,

Coordinates of the focus :

$(a,0)=(3,0)$

Axis of the parabola:

It can be seen that the axis of this parabola is X-Axis.

The equation of the directrix

$x=-a,\Rightarrow x=-3\Rightarrow x+3=0$

The length of the latus rectum:

$4a=4(3)=12$ .

Question:2 Find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum.

$x^2 = 6y$

Answer:

Given, a parabola with equation

$x^2 =6y$

This is parabola of the form $x^2=4ay$ which opens upward.

So,

By comparing the given parabola equation with the standard equation, we get,

$4a=6$

$a=\frac{3}{2}$

Hence,

Coordinates of the focus :

$(0,a)=\left (0,\frac{3}{2}\right)$

Axis of the parabola:

It can be seen that the axis of this parabola is Y-Axis.

The equation of the directrix

$y=-a,\Rightarrow y=-\frac{3}{2}\Rightarrow y+\frac{3}{2}=0$

The length of the latus rectum:

$4a=4(\frac{3}{2})=6$ .

Question:3 Find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum.

$y^2 = -8x$

Answer:

Given, a parabola with equation

$y^2 =-8x$

This is parabola of the form $y^2=-4ax$ which opens towards left.

So,

By comparing the given parabola equation with the standard equation, we get,

$-4a=-8$

$a=2$

Hence,

Coordinates of the focus :

$(-a,0)=(-2,0)$

Axis of the parabola:

It can be seen that the axis of this parabola is X-Axis.

The equation of the directrix

$x=a,\Rightarrow x=2\Rightarrow x-2=0$

The length of the latus rectum:

$4a=4(2)=8$ .

Question:4 Find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum.

$x^2 = -16y$

Answer:

Given, a parabola with equation

$x^2 =-16y$

This is parabola of the form $x^2=-4ay$ which opens downwards.

So,

By comparing the given parabola equation with the standard equation, we get,

$-4a=-16$

$a=4$

Hence,

Coordinates of the focus :

$(0,-a)=(0,-4)$

Axis of the parabola:

It can be seen that the axis of this parabola is Y-Axis.

The equation of the directrix

$y=a,\Rightarrow y=4\Rightarrow y-4=0$

The length of the latus rectum:

$4a=4(4)=16$ .

Question:5 Find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum.

$y^2 = 10x$

Answer:

Given, a parabola with equation

$y^2 =10x$

This is parabola of the form $y^2=4ax$ which opens towards the right.

So,

By comparing the given parabola equation with the standard equation, we get,

$4a=10$

$a=\frac{10}{4}=\frac{5}{2}$

Hence,

Coordinates of the focus :

$(a,0)=\left(\frac{5}{2},0\right)$

Axis of the parabola:

It can be seen that the axis of this parabola is X-Axis.

The equation of the directrix

$x=-a,\Rightarrow x=-\frac{5}{2}\Rightarrow x+\frac{5}{2}=0\Rightarrow 2x+5=0$

The length of the latus rectum:

$4a=4(\frac{5}{2})=10$ .

Question:6 Find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum.

$x^2 = -9y$

Answer:

Given, a parabola with equation

$x^2 =-9y$

This is parabola of the form $x^2=-4ay$ which opens downwards.

By comparing the given parabola equation with the standard equation, we get,

$-4a=-9$

$a=\frac{9}{4}$

Hence,

Coordinates of the focus :

$(0,-a)=\left (0,-\frac{9}{4}\right)$

Axis of the parabola:

It can be seen that the axis of this parabola is Y-Axis.

The equation of the directrix

$y=a,\Rightarrow y=\frac{9}{4}\Rightarrow y-\frac{9}{4}=0$

The length of the latus rectum:

$4a=4\left(\frac{9}{4}\right)=9$ .

Question:7 Find the equation of the parabola that satisfies the given conditions:

Focus (6,0); directrix $x = - 6$

Answer:

Given, in a parabola,

Focus : (6,0) And Directrix : $x = - 6$

Here,

Focus is of the form (a, 0), which means it lies on the X-axis. And Directrix is of the form $x=-a$ which means it lies left to the Y-Axis.

These are the condition when the standard equation of a parabola is. $y^2=4ax$

Hence the Equation of Parabola is

$y^2=4ax$

Here, it can be seen that:

$a=6$

Hence the Equation of the Parabola is:

$\Rightarrow y^2=4ax\Rightarrow y^2=4(6)x$

$\Rightarrow y^2=24x$ .

Question:8 Find the equation of the parabola that satisfies the given conditions:

Focus (0,–3); directrix $y = 3$

Answer:

Given,in a parabola,

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

Here,

Focus is of the form (0,-a), which means it lies on the Y-axis. And Directrix is of the form $y=a$ which means it lies above X-Axis.

These are the conditions when the standard equation of a parabola is $x^2=-4ay$ .

Hence the Equation of Parabola is

$x^2=-4ay$

Here, it can be seen that:

$a=3$

Hence the Equation of the Parabola is:

$\Rightarrow x^2=-4ay\Rightarrow x^2=-4(3)y$

$\Rightarrow x^2=-12y$ .

Question:9 Find the equation of the parabola that satisfies the given conditions:

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

Answer:

Given,

Vertex (0,0) And focus (3,0)

As vertex of the parabola is (0,0) and focus lies in the positive X-axis, The parabola will open towards the right, And the standard equation of such parabola is

$y^2=4ax$

Here it can be seen that $a=3$

So, the equation of a parabola is

$\Rightarrow y^2=4ax\Rightarrow y^2=4(3)x$

$\Rightarrow y^2=12x$ .

Question:10 Find the equation of the parabola that satisfies the given conditions:

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

Answer:

Given,

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

As vertex of the parabola is (0,0) and focus lies in the negative X-axis, The parabola will open towards left, And the standard equation of such parabola is

$y^2=-4ax$

Here it can be seen that $a=2$

So, the equation of a parabola is

$\Rightarrow y^2=-4ax\Rightarrow y^2=-4(2)x$

$\Rightarrow y^2=-8x$ .

Question:11 Find the equation of the parabola that satisfies the given conditions:

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

Answer:

Given

The Vertex of the parabola is (0,0).

The parabola is passing through (2,3) and axis is along the x-axis, it will open towards right. and the standard equation of such parabola is

$y^2=4ax$

Now since it passes through (2,3)

$3^2=4a(2)$

$9=8a$

$a=\frac{8}{9}$

So the Equation of Parabola is ;

$\Rightarrow y^2=4\left(\frac{9}{8}\right)x$

$\Rightarrow y^2=\left(\frac{9}{2}\right)x$

$\Rightarrow 2y^2=9x$

Question:12 Find the equation of the parabola that satisfies the given conditions:

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

Answer:

Given a parabola,

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

Since the parabola is symmetric with respect to Y=axis, it's axis will ve Y-axis. and since it passes through the point (5,2), it must go through the first quadrant.

So the standard equation of such parabola is

$x^2=4ay$

Now since this parabola is passing through (5,2)

$5^2=4a(2)$

$25=8a$

$a=\frac{25}{8}$

Hence the equation of the parabola is

$\Rightarrow x^2=4\left ( \frac{25}{8} \right )y$

$\Rightarrow x^2=\left ( \frac{25}{2} \right )y$

$\Rightarrow 2x^2=25y$

Benefits of NCERT Solutions for Class 11 Maths Chapter 11 Exercise 11.2:-

As many questions in the engineering entrance exam are asked from the parabola, exercise 11.2 Class 11 Maths becomes very important for the competitive exams.
You must try to solve all the NCERT problems of Class 11 Maths chapter 11 exercise 11.2 in order to get conceptual clarity.
Class 11 Maths chapter 11 exercise 11.2 solutions are here to help you when you are not able to solve NCERT problems by yourself.

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Frequently Asked Questions (FAQs)

1. When a ball is throwing in air in influence of gravity, the ball will follow the path ?

When a ball is throwing in the air in presence of gravity, the ball will follow the parabolic path.

2. Find the coordinates of the focus of the parabola y^2 = 8x .

Equation of the parabola => y^2 = 8x

Comparing with the given equation => y^2 = 4ax

a = 2

The coordinate of the focus of the parabola is (2, 0)

3. Find the coordinates of the equation of the directrix of the parabola y^2 = 8x .

Equation of the parabola => y^2 = 8x

Comparing with the given equation => y^2 = 4ax

a = 2

The equation of the directrix of the parabola is x = – 2

4. Find the length of the latus rectum of the parabola y^2 = 8x .

Equation of the parabola => y^2 = 8x

Comparing with the given equation => y^2 = 4ax

a = 2

Length of the latus rectum is (4a) = 4 x 2 = 8.

5. Write the equation of the with focus at (a, 0) directrix x = – a (a > 0)

The equation of the parabola with focus at (a, 0) and directrix x = – a is y^2 = 4ax.

6. What is the length of lotus rectum of the parabola y^2 = 4ax.

The length of the latus rectum of the parabola y^2 = 4ax is 4a.

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