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.

JEE Main Scholarship Test Kit (Class 11): Narayana | Physics Wallah | Aakash | Unacademy

Suggested: JEE Main: high scoring chapters | Past 10 year's papers

Scholarship Test: Vidyamandir Intellect Quest (VIQ)

This Story also Contains

Conic Section Class 11 Chapter 11 Exercise 11.2
More About NCERT Solutions for Class 11 Maths Chapter 11 Exercise 11.2:-
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.

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 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.

Also Read

NCERT Solutions for Class 11 Physics Chapter 6 Work Energy and Power

NCERT Solutions for Class 11 Chemistry Chapter 5 States of Matter

NCERT Class 11 Physics Chapter 5 Notes Laws of Motion - Download PDF

NCERT Class 11 Physics Chapter 3 Notes Motion in a straight Line - Download PDF

NCERT Solutions for Exercise 10.1 Class 11 Maths Chapter 10 - Straight Lines

NCERT Solutions for Miscellaneous Exercise Chapter 16 Class 11 - Probability

NCERT Class 11 Chemistry Chapter 4 Notes Chemical Bonding and Molecular Structure - Download PDF

NCERT Class 11 Chemistry Chapter 3 Notes - Download PDF

NCERT Class 11 Biology Chapter 19 Notes Excretory Products And Their Elimination- Download PDF Notes

NCERT Class 11 Biology Chapter 22 Notes Chemical Coordination And Integration- Download PDF Notes

Some basic concepts of Chemistry Class 11th Notes - Free NCERT Class 11 Chemistry Chapter 1 Notes - Download PDF

NCERT Class 11 Biology Chapter 21 Notes Neural Control And Coordination- Download PDF Notes

NCERT Exemplar Class 11 Maths Solutions Chapter 12 Introduction to Three Dimensional Geometry

NCERT Exemplar Class 11 Maths Solutions Chapter 10 Straight Lines

NCERT Exemplar Class 11 Maths Solutions Chapter 6 Linear Inequalities

NCERT Exemplar Class 11 Maths Solutions Chapter 16 Probability

NCERT Exemplar Class 11 Maths Solutions Chapter 14 Mathematical Reasoning

NCERT Exemplar Class 11 Maths Solutions Chapter 7 Permutations and Combinations

Articles

Silverzone ABHO Result 2024-25 - Check Akhil Bhartiya Hindi Olympiad Results Here

Nov 20, 2024

Silverzone SKGKO Result 2024-25 – Check Your Scores and Performance

Nov 20, 2024

Education Loan for School Students in India – Eligibility, Benefits & Application Process

Nov 20, 2024

Silverzone ABHO Answer Key 2024 | Download Akhil Bhartiya Hindi Olympiad Solutions

Nov 20, 2024

Silverzone SKGKO Answer Key 2024-25: Download Smart Kid GK Olympiad Answer Key @ silverzone.org

Nov 20, 2024

Sainik School Entrance Exam 2025: Application, Exam Date, How to Apply, Syllabus

Nov 20, 2024

ICICI Education Loan: Eligibility, Interest Rates & Application Process

Nov 19, 2024

MPSOS 10th Result 2024 - Check MPSOS Class 10 December 2024 Exam Result @mpsos.nic.in

Nov 16, 2024

Upcoming School Exams

National Institute of Open Schooling 12th Examination

Admit Card Date:04 October,2024 - 29 November,2024

Application Process

Exam Pattern

Admit Card

National Institute of Open Schooling 10th examination

Admit Card Date:04 October,2024 - 29 November,2024

Application Process

Exam Pattern

Mock Test

Mizoram Board of School Education 12th Examination

Application Date:07 October,2024 - 22 November,2024

Exam Pattern

Result

Mock Test

Mizoram Board 10th Examination

Application Date:07 October,2024 - 22 November,2024

Exam Pattern

Admit Card

Result

Punjab Board of Secondary Education 10th Examination

Application Correction Date:08 October,2024 - 27 November,2024

Exam Pattern

Result

Admit Card

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) $0.67\; J$

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 3.8×10⁷J 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) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

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) $K/4$

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 .

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

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