Careers360 Logo
School
Search Colleges, Exams, Schools & more
NCERT Solutions for Exercise 11.4 Class 11 Maths Chapter 11 - Conic Section

NCERT Solutions for Exercise 11.4 Class 11 Maths Chapter 11 - Conic Section

Edited By Ravindra Pindel | Updated on Jul 18, 2022 11:54 AM IST

In the previous exercises of this chapter, you have already learned about circles, parabola, and ellipse. In the NCERT solutions for Class 11 Maths chapter 11 exercise 11.4, you will learn about another conic section called 'Hyperbola'. The set of all points in a plane, the difference of whose distances from two fixed points in the plane is a constant is called a hyperbola. In the real world like open orbits of some comets about the sun follow hyperbolas. Hyperbola has applications in the field of science, research, and design in the design of bridges.

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
  1. Conic Section Class 11 Chapter 11 Exercise 11.4
  2. More About NCERT Solutions for Class 11 Maths Chapter 11 Exercise 11.4:-
  3. Benefits of NCERT Solutions for Class 11 Maths Chapter 11 Exercise 11.4:-
  4. NCERT Solutions of Class 11 Subject Wise
  5. Subject Wise NCERT Exampler Solutions

The eccentricity, standard equation of hyperbola, latus rectum are also covered in the Class 11 Maths chapter 11 exercise 11.4 solution. You are advised to go through the theories, definitions, and observations given in the NCERT textbook before exercise 11.4 Class 11 Maths. After that, you can try to solve the given solved examples and NCERT syllabus problems by yourself. You can Class 11 Maths chapter 11 exercise 11.4 solutions whenever find difficulty while solving them. Click on the NCERT Solutions link if you are looking for NCERT solutions at one place.

Also, see

Conic Section Class 11 Chapter 11 Exercise 11.4

Question:1 Find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbolas.

\frac{x^2}{16} - \frac{y^2}{9} = 1

Answer:

Given a Hyperbola equation,

\frac{x^2}{16} - \frac{y^2}{9} = 1

Can also be written as

\frac{x^2}{4^2} - \frac{y^2}{3^2} = 1

Comparing this equation with the standard equation of the hyperbola:

\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1

We get,

a=4 and b=3

Now, As we know the relation in a hyperbola,

c^2=a^2+b^2

c=\sqrt{a^2+b^2}

c=\sqrt{4^2+3^2}

c=5

Here as we can see from the equation that the axis of the hyperbola is X -axis. So,

Coordinates of the foci:

(c,0) \:and\:(-c,0)=(5,0)\:and\:(-5,0)

The Coordinates of vertices:

(a,0) \:and\:(-a,0)=(4,0)\:and\:(-4,0)

The Eccentricity:

e=\frac{c}{a}=\frac{5}{4}

The Length of the latus rectum :

\frac{2b^2}{a}=\frac{2(3)^2}{4}=\frac{18}{4}=\frac{9}{2}

Question:2Find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbolas.

\frac{y^2}{9} - \frac{x^2}{27} = 1

Answer:

Given a Hyperbola equation,

\frac{y^2}{9} - \frac{x^2}{27} = 1

Can also be written as

\frac{y^2}{3^2} - \frac{x^2}{(\sqrt{27})^2} = 1

Comparing this equation with the standard equation of the hyperbola:

\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1

We get,

a=3 and b=\sqrt{27}

Now, As we know the relation in a hyperbola,

c^2=a^2+b^2

c=\sqrt{a^2+b^2}

c=\sqrt{3^2+(\sqrt{27})^2}

c=\sqrt{36}

c=6

Here as we can see from the equation that the axis of the hyperbola is Y-axis. So,

Coordinates of the foci:

(0,c) \:and\:(0,-c)=(0,6)\:and\:(0,-6)

The Coordinates of vertices:

(0,a) \:and\:(0,-a)=(0,3)\:and\:(0,-3)

The Eccentricity:

e=\frac{c}{a}=\frac{6}{3}=2

The Length of the latus rectum :

\frac{2b^2}{a}=\frac{2(27)}{3}=\frac{54}{3}=18

Question:3 Find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbolas.

9 y^2 - 4 x^2 =36

Answer:

Given a Hyperbola equation,

9 y^2 - 4 x^2 =36

Can also be written as

\frac{9y^2}{36} - \frac{4x^2}{36} = 1

\frac{y^2}{2^2} - \frac{x^2}{3^2} = 1

Comparing this equation with the standard equation of the hyperbola:

\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1

We get,

a=2 and b=3

Now, As we know the relation in a hyperbola,

c^2=a^2+b^2

c=\sqrt{a^2+b^2}

c=\sqrt{2^2+3^2}

c=\sqrt{13}

Hence,

Coordinates of the foci:

(0,c) \:and\:(0,-c)=(0,\sqrt{13})\:and\:(0,-\sqrt{13})

The Coordinates of vertices:

(0,a) \:and\:(0,-a)=(0,2)\:and\:(0,-2)

The Eccentricity:

e=\frac{c}{a}=\frac{\sqrt{13}}{2}

The Length of the latus rectum :

\frac{2b^2}{a}=\frac{2(9)}{2}=\frac{18}{2}=9

Question:4 Find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbolas.

16x^2 - 9y^2 = 576

Answer:

Given a Hyperbola equation,

16x^2 - 9y^2 = 576

Can also be written as

\frac{16x^2}{576} - \frac{9y^2}{576} = 1

\frac{x^2}{36} - \frac{y^2}{64} = 1

\frac{x^2}{6^2} - \frac{y^2}{8^2} = 1

Comparing this equation with the standard equation of the hyperbola:

\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1

We get,

a=6 and b=8

Now, As we know the relation in a hyperbola,

c^2=a^2+b^2

c=\sqrt{a^2+b^2}

c=\sqrt{6^2+8^2}

c=10

Therefore,

Coordinates of the foci:

(c,0) \:and\:(-c,0)=(10,0)\:and\:(-10,0)

The Coordinates of vertices:

(a,0) \:and\:(-a,0)=(6,0)\:and\:(-6,0)

The Eccentricity:

e=\frac{c}{a}=\frac{10}{6}=\frac{5}{3}

The Length of the latus rectum :

\frac{2b^2}{a}=\frac{2(8)^2}{6}=\frac{128}{6}=\frac{64}{3}

Question:5 Find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbolas.

5y^2 - 9x^2 = 36

Answer:

Given a Hyperbola equation,

5y^2 - 9x^2 = 36

Can also be written as

\frac{5y^2}{36} - \frac{9x^2}{36} = 1

\frac{y^2}{\frac{36}{5}} - \frac{x^2}{4} = 1

\frac{y^2}{(\frac{6}{\sqrt{5}})^2} - \frac{x^2}{2^2} = 1

Comparing this equation with the standard equation of the hyperbola:

\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1

We get,

a=\frac{6}{\sqrt{5}}

and b=2

Now, As we know the relation in a hyperbola,

c^2=a^2+b^2

c=\sqrt{a^2+b^2}

c=\sqrt{(\frac{6}{\sqrt{5}})^2+2^2}

c=\sqrt{\frac{56}{5}}

c=2\sqrt{\frac{14}{5}}

Here as we can see from the equation that the axis of the hyperbola is Y-axis. So,

Coordinates of the foci:

(0,c) \:and\:(0,-c)=\left(0,2\sqrt{\frac{14}{5}}\right)\:and\:\left(0,-2\sqrt{\frac{14}{5}}\right)

The Coordinates of vertices:

(0,a) \:and\:(0,-a)=\left(0,\frac{6}{\sqrt{5}}\right)\:and\:\left(0,-\frac{6}{\sqrt{5}}\right)

The Eccentricity:

e=\frac{c}{a}=\frac{2\sqrt{\frac{14}{5}}}{\frac{6}{\sqrt{5}}}=\frac{\sqrt{14}}{3}

The Length of the latus rectum :

\frac{2b^2}{a}=\frac{2(4)}{\frac{6}{\sqrt{5}}}=\frac{4\sqrt{5}}{3}

Question:6 Find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbolas.

49y^2 - 16x^2 = 784

Answer:

Given a Hyperbola equation,

49y^2 - 16x^2 = 784

Can also be written as

\frac{49y^2}{784} - \frac{16x^2}{784} = 1

\frac{y^2}{16} - \frac{x^2}{49} = 1

\frac{y^2}{4^2} - \frac{x^2}{7^2} = 1

Comparing this equation with the standard equation of the hyperbola:

\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1

We get,

a=4 and b=7

Now, As we know the relation in a hyperbola,

c^2=a^2+b^2

c=\sqrt{a^2+b^2}

c=\sqrt{4^2+7^2}

c=\sqrt{65}

Therefore,

Coordinates of the foci:

(0,c) \:and\:(0,-c)=(0,\sqrt{65})\:and\:(0,-\sqrt{65})

The Coordinates of vertices:

(0,a) \:and\:(0,-a)=(0,4)\:and\:(0,-4)

The Eccentricity:

e=\frac{c}{a}=\frac{\sqrt{65}}{4}

The Length of the latus rectum :

\frac{2b^2}{a}=\frac{2(49)}{4}=\frac{98}{4}=\frac{49}{2}

Question:7 Find the equations of the hyperbola satisfying the given conditions.

Vertices (± 2, 0), foci (± 3, 0)

Answer:

Given, in a hyperbola

Vertices (± 2, 0), foci (± 3, 0)

Here, Vertices and focii are on the X-axis so, the standard equation of the Hyperbola will be ;

\frac{x^2}{a^2}-\frac{y^2}{b^2}=1

By comparing the standard parameter (Vertices and foci) with the given one, we get

a=2 and c=3

Now, As we know the relation in a hyperbola

c^2=a^2+b^2

b^2=c^2-a^2

b^2=3^2-2^2

b^2=9-4=5

Hence,The Equation of the hyperbola is ;

\frac{x^2}{4}-\frac{y^2}{5}=1

Question:8 Find the equations of the hyperbola satisfying the given conditions.

Vertices (0, ± 5), foci (0, ± 8)

Answer:

Given, in a hyperbola

Vertices (0, ± 5), foci (0, ± 8)

Here, Vertices and focii are on the Y-axis so, the standard equation of the Hyperbola will be ;

\frac{y^2}{a^2}-\frac{x^2}{b^2}=1

By comparing the standard parameter (Vertices and foci) with the given one, we get

a=5 and c=8

Now, As we know the relation in a hyperbola

c^2=a^2+b^2

b^2=c^2-a^2

b^2=8^2-5^2

b^2=64-25=39

Hence, The Equation of the hyperbola is ;

\frac{y^2}{25}-\frac{x^2}{39}=1.

Question:9 Find the equations of the hyperbola satisfying the given conditions.

Vertices (0, ± 3), foci (0, ± 5)

Answer:

Given, in a hyperbola

Vertices (0, ± 3), foci (0, ± 5)

Here, Vertices and focii are on the Y-axis so, the standard equation of the Hyperbola will be ;

\frac{y^2}{a^2}-\frac{x^2}{b^2}=1

By comparing the standard parameter (Vertices and foci) with the given one, we get

a=3 and c=5

Now, As we know the relation in a hyperbola

c^2=a^2+b^2

b^2=c^2-a^2

b^2=5^2-3^2

b^2=25-9=16

Hence, The Equation of the hyperbola is ;

\frac{y^2}{9}-\frac{x^2}{16}=1.

Question:10 Find the equations of the hyperbola satisfying the given conditions.

Foci (± 5, 0), the transverse axis is of length 8.

Answer:

Given, in a hyperbola

Foci (± 5, 0), the transverse axis is of length 8.

Here, focii are on the X-axis so, the standard equation of the Hyperbola will be ;

\frac{x^2}{a^2}-\frac{y^2}{b^2}=1

By comparing the standard parameter (transverse axis length and foci) with the given one, we get

2a=8\Rightarrow a=4 and c=5

Now, As we know the relation in a hyperbola

c^2=a^2+b^2

b^2=c^2-a^2

b^2=5^2-4^2

b^2=25-16=9

Hence, The Equation of the hyperbola is ;

\frac{x^2}{16}-\frac{y^2}{9}=1

Question:11 Find the equations of the hyperbola satisfying the given conditions.

Foci (0, ±13), the conjugate axis is of length 24.

Answer:

Given, in a hyperbola

Foci (0, ±13), the conjugate axis is of length 24.

Here, focii are on the Y-axis so, the standard equation of the Hyperbola will be ;

\frac{y^2}{a^2}-\frac{x^2}{b^2}=1

By comparing the standard parameter (length of conjugate axis and foci) with the given one, we get

2b=24\Rightarrow b=12 and c=13

Now, As we know the relation in a hyperbola

c^2=a^2+b^2

a^2=c^2-b^2

a^2=13^2-12^2

a^2=169-144=25

Hence, The Equation of the hyperbola is ;

\frac{y^2}{25}-\frac{x^2}{144}=1.

Question:12 Find the equations of the hyperbola satisfying the given conditions.

Foci (\pm 3\sqrt5, 0), the latus rectum is of length 8.

Answer:

Given, in a hyperbola

Foci (\pm 3\sqrt5, 0), the latus rectum is of length 8.

Here, focii are on the X-axis so, the standard equation of the Hyperbola will be ;

\frac{x^2}{a^2}-\frac{y^2}{b^2}=1

By comparing standard parameter (length of latus rectum and foci) with the given one, we get

c=3\sqrt{5} and

\frac{2b^2}{a}=8\Rightarrow 2b^2=8a\Rightarrow b^2=4a

Now, As we know the relation in a hyperbola

c^2=a^2+b^2

c^2=a^2+4a

a^2+4a=(3\sqrt{5})^2

a^2+4a=45

a^2+9a-5a-45=0

(a+9)(a-5)=0

a=-9\:or\:5

Since a can never be negative,

a=5

a^2=25

b^2=4a=4(5)=20

Hence, The Equation of the hyperbola is ;

\frac{x^2}{25}-\frac{y^2}{20}=1

Question:13 Find the equations of the hyperbola satisfying the given conditions.

Foci (± 4, 0), the latus rectum is of length 12

Answer:

Given, in a hyperbola

Foci (± 4, 0), the latus rectum is of length 12

Here, focii are on the X-axis so, the standard equation of the Hyperbola will be ;

\frac{x^2}{a^2}-\frac{y^2}{b^2}=1

By comparing standard parameter (length of latus rectum and foci) with the given one, we get

c=4 and

\frac{2b^2}{a}=12\Rightarrow 2b^2=12a\Rightarrow b^2=6a

Now, As we know the relation in a hyperbola

c^2=a^2+b^2

c^2=a^2+6a

a^2+6a=4^2

a^2+6a=16

a^2+8a-2a-16=0

(a+8)(a-2)=0

a=-8\:or\:2

Since a can never be negative,

a=2

a^2=4

b^2=6a=6(2)=12

Hence, The Equation of the hyperbola is ;

\frac{x^2}{4}-\frac{y^2}{12}=1

Question:14 Find the equations of the hyperbola satisfying the given conditions.

vertices (± 7,0), e = \frac{4}{3}

Answer:

Given, in a hyperbola

vertices (± 7,0), And

e = \frac{4}{3}

Here, Vertices is on the X-axis so, the standard equation of the Hyperbola will be ;

\frac{x^2}{a^2}-\frac{y^2}{b^2}=1

By comparing the standard parameter (Vertices and eccentricity) with the given one, we get

a=7 and

e=\frac{c}{a}=\frac{c}{7}=\frac{4}{3}

From here,

c=\frac{28}{3}

Now, As we know the relation in a hyperbola

c^2=a^2+b^2

b^2=c^2-a^2

b^2=\left(\frac{28}{3}\right)^2-7^2

b^2=\left(\frac{784}{9}\right)-49

b^2=\left(\frac{784-441}{9}\right)=\frac{343}{9}

Hence, The Equation of the hyperbola is ;

\frac{x^2}{49}-\frac{9y^2}{343}=1

Question:15 Find the equations of the hyperbola satisfying the given conditions.

Foci (0,\pm\sqrt{10}), passing through (2,3)

Answer:

Given, in a hyperbola,

Foci (0,\pm\sqrt{10}), passing through (2,3)

Since foci of the hyperbola are in Y-axis, the equation of the hyperbola will be of the form ;

\frac{y^2}{a^2}-\frac{x^2}{b^2}=1

By comparing standard parameter (foci) with the given one, we get

c=\sqrt{10}

Now As we know, in a hyperbola

a^2+b^2=c^2

a^2+b^2=10\:\:\:\:\:\:\:....(1)

Now As the hyperbola passes through the point (2,3)

\frac{3^2}{a^2}-\frac{2^2}{b^2}=1

9b^2-4a^2=a^2b^2\:\;\;\:\:\;\:....(2)

Solving Equation (1) and (2)

9(10-a^2)-4a^2=a^2(10-a^2)

a^4-23a^2+90=0

(a^2)^2-18a^2-5a^2+90=0

(a^2-18)(a^2-5)=0

a^2=18\:or\:5

Now, as we know that in a hyperbola c is always greater than, a we choose the value

a^2=5

b^2=10-a^2=10-5=5

Hence The Equation of the hyperbola is

\frac{y^2}{5}-\frac{x^2}{5}=1

More About NCERT Solutions for Class 11 Maths Chapter 11 Exercise 11.4:-

Class 11 Maths chapter 11 exercise 11.4 consists of questions related to finding the coordinates of the foci of the hyperbola, vertices of the hyperbola, the eccentricity of the hyperbola, and the length of the latus rectum of the hyperbola. Also, you will get questions related to writing the equation of hyperbola where the other parameters of the hyperbola are given.

Also Read| Conic Section Class 11 Notes

Benefits of NCERT Solutions for Class 11 Maths Chapter 11 Exercise 11.4:-

  • NCERT solutions for Class 11 Maths chapter 11 exercise 11.4 are designed by subject matter experts who are experienced in this field.
  • Class 11 Maths chapter 11 exercise 11.4 solutions are explained in a very detailed manner which could be understood by an average student also.
  • You can use these Class 11 Maths chapter 11 exercise 11.4 solutions to get help in your homework.

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. What is the definition of hyperbola ?

The set of all points in a plane, the difference of whose distances from two fixed points in the plane is a constant is called a hyperbola.

2. What is the centre of the hyperbola ?

The center of the hyperbola is the mid-point of the line segment joining the foci.

3. What is the transverse axis of the hyperbola ?

The transverse axis is the line passing through the foci of the hyperbola.

4. What is the conjugate axis of the hyperbola ?

The conjugate axis is called the line passing through the centre of the hyperbola and perpendicular to the transverse axis.

5. Can the eccentricity of hyperbola be less than one ?

No, the eccentricity of a hyperbola is always greater than one.

6. what is the eccentricity of parabola ?

The eccentricity of a parabola is always one.

Also Read

Articles

Upcoming School Exams

Himachal Pradesh Board of Secondary Education 10th Examination

Application Date:09 September,2024 - 14 November,2024

Himachal Pradesh Board of Secondary Education 12th Examination

Application Date:09 September,2024 - 14 November,2024

National Institute of Open Schooling 12th Examination

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

National Institute of Open Schooling 10th examination

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

National Means Cum-Merit Scholarship

Application Date:05 October,2024 - 04 November,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

Applications for Admissions are open.

VMC VIQ Scholarship Test
VMC VIQ Scholarship Test
Apply

Register for Vidyamandir Intellect Quest. Get Scholarship and Cash Rewards.

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

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

TOEFL ® Registrations 2024
TOEFL ® Registrations 2024
Apply

Accepted by more than 11,000 universities in over 150 countries worldwide

Pearson | PTE
Pearson | PTE
Apply

Register now for PTE & Unlock 20% OFF : Use promo code: 'C360SPL20'. Valid till 15th NOV'24! Trusted by 3,500+ universities globally

JEE Main high scoring chapters and topics
JEE Main high scoring chapters and topics
Apply

As per latest 2024 syllabus. Study 40% syllabus and score upto 100% marks in JEE

View All Application Forms
News and Notifications
CM Atishi, Sisodia inaugurate school building in east Delhi
Class 6 student dies in UP residential school
15-year-old girl sexually harassed by schoolmates in Uttar Pradesh
Saroj Sharma not removed as chairperson, repatriated as per her request, says NIOS
Education News This Week: CBSE practical exam dates; IIT Delhi suicide; IIT Bombay’s facelift
NHRC takes suo motu cognizance of Class 8 student’s alleged suicide in private school at Guntur
UP: Masked men open fire on school van, no injuries reported
Back to top