Careers360 Logo
School
Search Colleges, Exams, Schools & more
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.

This Story also Contains
  1. Conic Section Class 11 Chapter 11 Exercise 11.2
  2. More About NCERT Solutions for Class 11 Maths Chapter 11 Exercise 11.2:-
  3. Benefits of NCERT Solutions for Class 11 Maths Chapter 11 Exercise 11.2:-
  4. NCERT Solutions of Class 11 Subject Wise
  5. NCERT Solutions for Class 11 Maths
  6. 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.

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.

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

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%
Download EBook

NCERT Solutions of Class 11 Subject Wise

Subject Wise NCERT Exampler Solutions

Happy learning!!!

Frequently Asked Question (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.

Also Read

Articles

Upcoming School Exams

National Institute of Open Schooling 12th Examination

Admit Card Date:28 March,2024 - 22 May,2024

National Institute of Open Schooling 10th examination

Admit Card Date:28 March,2024 - 22 May,2024

Bihar Board 12th Examination

Admit Card Date:19 April,2024 - 11 May,2024

Karnataka Pre-University Certificate Examination

Exam Date:03 May,2024 - 03 May,2024

Bihar Board 10th Examination

Exam Date:04 May,2024 - 11 May,2024

View All School Exams
Get answers from students and experts

Popular Questions

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)

0.34\; J

 Option 2)

0.16\; J
Option 3)

1.00\; J

 Option 4)

0.67\; J
Read More

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

 
Read More

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)

2,000 \; J - 5,000\; J

 Option 2)

200 \, \, J - 500 \, \, J
Option 3)

2\times 10^{5}J-3\times 10^{5}J

 Option 4)

20,000 \, \, J - 50,000 \, \, J
Read More

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

Option 1)

K/2\,

 Option 2)

\; K\;
Option 3)

zero\;

 Option 4)

K/4
Read More

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

 Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced
Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

 Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .
Read More

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} 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
Read More

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.
Read More

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.
Read More

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
Read More

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
Read More

Colleges After 12th

Popular Course After 12th

Also Read

Applications for Admissions are open.

JEE Main Important Physics formulas
JEE Main Important Physics formulas
Apply

As per latest 2024 syllabus. Physics formulas, equations, & laws of class 11 & 12th chapters

UPES School of Liberal Studies
UPES School of Liberal Studies
Apply

Ranked #52 Among Universities in India by NIRF | Up to 30% Merit-based Scholarships | Lifetime placement assistance

Aakash iACST Scholarship Test 2024
Aakash iACST Scholarship Test 2024
Apply

Get up to 90% scholarship on NEET, JEE & Foundation courses

JEE Main Important Chemistry formulas
JEE Main Important Chemistry formulas
Apply

As per latest 2024 syllabus. Chemistry formulas, equations, & laws of class 11 & 12th chapters

PACE IIT & Medical, Financial District, Hyd
PACE IIT & Medical, Financial District, Hyd
Apply

Enrol in PACE IIT & Medical, Financial District, Hyd for JEE/NEET preparation

ALLEN JEE Exam Prep
ALLEN JEE Exam Prep
Apply

Start your JEE preparation with ALLEN

View All Application Forms
News and Notifications
Barmer teacher suspended for slapping class 3 student
01/05/2024, 17:07:10
Teacher arrested for `sexual exploitation' at Ujjain Ashram school
01/05/2024, 14:28:18
Delhi NCR schools get bomb threat; fire tender arrives, search underway
01/05/2024, 04:09:04
Education is greatest gift to society: Vice President Jagdeep Dhankhar
30/04/2024, 17:02:12
Jharkhand schools up to Class 8 closed till further orders due to heat wave
30/04/2024, 11:49:37
Delhi School EWS Admission 2024-25: Registration window opens at edudel.nic.in; eligibility
30/04/2024, 09:27:25
Education ministry asks NCERT to review, update textbooks on yearly basis
29/04/2024, 12:22:03
Back to top