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

# 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.

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

$y^2 =12x$

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:

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

$x^2 = 6y$

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:

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

$y^2 = -8x$

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:

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

$x^2 = -16y$

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:

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

$y^2 = 10x$

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:

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

$x^2 = -9y$

Given, a parabola with equation

$x^2 =-9y$

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,

$-4a=-9$

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

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:

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

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

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:

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

$\Rightarrow y^2=24x$.

Focus (0,–3); directrix $y = 3$

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:

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

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

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

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

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

$\Rightarrow y^2=12x$.

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

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

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

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

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

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 ;

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

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

$\Rightarrow 2y^2=9x$

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

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

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

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

$\Rightarrow 2x^2=25y$

## More About NCERT Solutions for Class 11 Maths Chapter 11 Exercise 11.2:-

Class 11th Maths chapter 11 exercise 11.2 consists of questions related to coordinates of the focus of a parabola, the axis of the parabola, the equation of the directrix, and the length of the latus rectum given the equation of the parabola. There are some solved examples and theories related to parabola given before the Class 11 Maths chapter 11 exercise 11.2. You must go through the definitions and observations given in the NCERT textbook before solving the exercise 11.2 Class 11 Maths.

Also Read| Conic Section Class 11 Notes

## 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.

### Also see-

JEE Main Highest Scoring Chapters & Topics
Just Study 40% Syllabus and Score upto 100%

## Subject Wise NCERT Exampler Solutions

Happy learning!!!

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.

## Upcoming School Exams

#### National Means Cum-Merit Scholarship

Application Date:01 August,2024 - 16 September,2024

#### National Rural Talent Scholarship Examination

Application Date:05 September,2024 - 20 September,2024

Exam Date:19 September,2024 - 19 September,2024

Exam Date:20 September,2024 - 20 September,2024

#### National Institute of Open Schooling 12th Examination

Exam Date:20 September,2024 - 07 October,2024

Get answers from students and experts

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

 Option 1) Option 2) Option 3) Option 4)

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

 Option 1) 2.45×10−3 kg Option 2)  6.45×10−3 kg Option 3)  9.89×10−3 kg Option 4) 12.89×10−3 kg

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

 Option 1) Option 2) Option 3) Option 4)

A particle is projected at 600   to the horizontal with a kinetic energy . The kinetic energy at the highest point

 Option 1) Option 2) Option 3) Option 4)

In the reaction,

 Option 1)   at STP  is produced for every mole   consumed Option 2)   is consumed for ever      produced Option 3) is produced regardless of temperature and pressure for every mole Al that reacts Option 4) at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, will contain 0.25 mole of oxygen atoms?

 Option 1) 0.02 Option 2) 3.125 × 10-2 Option 3) 1.25 × 10-2 Option 4) 2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

 Option 1) decrease twice Option 2) increase two fold Option 3) remain unchanged Option 4) be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

 Option 1) Molality Option 2) Weight fraction of solute Option 3) Fraction of solute present in water Option 4) Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

 Option 1) twice that in 60 g carbon Option 2) 6.023 × 1022 Option 3) half that in 8 g He Option 4) 558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

 Option 1) less than 3 Option 2) more than 3 but less than 6 Option 3) more than 6 but less than 9 Option 4) more than 9