• School
  • Engineering and Architecture
  • Management and Business Administration
  • Medicine and Allied Sciences
  • Law
  • Animation and Design
  • Media, Mass Communication and Journalism
  • Finance & Accounts
  • Computer Application and IT
  • Pharmacy
  • Hospitality and Tourism
  • Competition
  • Study Abroad
  • Arts, Commerce & Sciences
  • Learn
  • Online Courses and Certifications
NCERT Solutions for Exercise 11.3 Class 11 Maths Chapter 11 - Conic Section

NCERT Solutions for Exercise 11.3 Class 11 Maths Chapter 11 - Conic Section

Edited By Ravindra Pindel | Updated on Jul 18, 2022 11:50 AM IST

In the previous exercises, you have already learned about the two conic sections called 'circle' and 'parabola'. In the NCERT solutions for Class 11 Maths chapter 11 exercise 11.3, you will learn about another conic section called 'Ellipse'. An ellipse is the set of all points in a plane, the sum of whose distances from two fixed points in the plane is a constant. There are two axes of the ellipse called the major axis and minor axis. In the Class 11 Maths chapter 11 exercise 11.3, you will also learn about the semi-minor axis, semi-major axis, and the relationship between the semi-minor axis and semi-major axis.

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: 11.3
  2. More About NCERT Solutions for Class 11 Maths Chapter 11 Exercise 11.3:-
  3. Benefits of NCERT Solutions for Class 11 Maths Chapter 11 Exercise 11.3:-
  4. NCERT Solutions of Class 11 Subject Wise
  5. Subject Wise NCERT Exampler Solutions

Some special cases of an ellipse, the eccentricity of an ellipse, standard equations of an ellipse, the latus rectum of an ellipse are also covered in the Class 11 Maths chapter 11 exercise 11.3 solutions. Many real-world situations like orbits of planets, orbits of satellites, orbits of moons, some airplane wings are represented by an ellipse. It has a lot of applications in the field of science, research, and design. There are some definitions, theories, and observations related to the ellipse is given in the NCERT textbook before this exercise. You must go through these theories in order to get conceptual clarity. In the Class 11th Maths chapter 11 exercise 11.3, you will learn about the mathematical applications of an ellipse. Also, you can check NCERT Solutions link if you are looking for NCERT solutions at one place.

Also, see

Conic Section Class 11 Chapter 11 Exercise: 11.3

Question:1 Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse.

\frac{x^2}{36} + \frac{y^2}{16} = 1

Answer:

Given

The equation of the ellipse

\frac{x^2}{36} + \frac{y^2}{16} = 1

As we can see from the equation, the major axis is along X-axis and the minor axis is along Y-axis.

On comparing the given equation with the standard equation of an ellipse, which is

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

We get

a=6 and b=4.

So,

c=\sqrt{a^2-b^2}=\sqrt{6^2-4^2}

c=\sqrt{20}=2\sqrt{5}

Hence,

Coordinates of the foci:

(c,0)\:and\:(-c,0)=(2\sqrt{5},0)\:and\:(-2\sqrt{5},0)

The vertices:

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

The length of the major axis:

2a=2(6)=12

The length of minor axis:

2b=2(4)=8

The eccentricity :

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

The length of the latus rectum:

\frac{2b^2}{a}=\frac{2(4)^2}{6}=\frac{32}{6}=\frac{16}{3}

Question:2 Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse.

\frac{x^2}{4} + \frac{y^2}{25} =1

Answer:

Given

The equation of the ellipse

\frac{x^2}{4} + \frac{y^2}{25} =1

As we can see from the equation, the major axis is along Y-axis and the minor axis is along X-axis.

On comparing the given equation with the standard equation of such ellipse, which is

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

We get

a=5 and b=2.

So,

c=\sqrt{a^2-b^2}=\sqrt{5^2-2^2}

c=\sqrt{21}

Hence,

Coordinates of the foci:

(0,c)\:and\:(0,-c)=(0,\sqrt{21})\:and\:(0,-\sqrt{21})

The vertices:

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

The length of the major axis:

2a=2(5)=10

The length of minor axis:

2b=2(2)=4

The eccentricity :

e=\frac{c}{a}=\frac{\sqrt{21}}{6}

The length of the latus rectum:

\frac{2b^2}{a}=\frac{2(2)^2}{5}=\frac{8}{5}

Question:3 Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse.

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

Answer:

Given

The equation of the ellipse

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

As we can see from the equation, the major axis is along X-axis and the minor axis is along Y-axis.

On comparing the given equation with the standard equation of an ellipse, which is

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

We get

a=4 and b=3.

So,

c=\sqrt{a^2-b^2}=\sqrt{4^2-3^2}

c=\sqrt{7}

Hence,

Coordinates of the foci:

(c,0)\:and\:(-c,0)=(\sqrt{7},0)\:and\:(-\sqrt{7},0)

The vertices:

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

The length of the major axis:

2a=2(4)=8

The length of minor axis:

2b=2(3)=6

The eccentricity :

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

The length of the latus rectum:

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

Question:4 Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse.

\frac{x^2}{25} + \frac{y^2}{100} = 1

Answer:

Given

The equation of the ellipse

\frac{x^2}{25} + \frac{y^2}{100} = 1

As we can see from the equation, the major axis is along Y-axis and the minor axis is along X-axis.

On comparing the given equation with the standard equation of such ellipse, which is

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

We get

a=10 and b=5.

So,

c=\sqrt{a^2-b^2}=\sqrt{10^2-5^2}

c=\sqrt{75}=5\sqrt{3}

Hence,

Coordinates of the foci:

(0,c)\:and\:(0,-c)=(0,5\sqrt{3})\:and\:(0,-5\sqrt{3})

The vertices:

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

The length of the major axis:

2a=2(10)=20

The length of minor axis:

2b=2(5)=10

The eccentricity :

e=\frac{c}{a}=\frac{5\sqrt{3}}{10}=\frac{\sqrt{3}}{2}

The length of the latus rectum:

\frac{2b^2}{a}=\frac{2(5)^2}{10}=\frac{50}{10}=5

Question:5 Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse.

\frac{x^2}{49} + \frac{y^2}{36} = 1

Answer:

Given

The equation of ellipse

\frac{x^2}{49} + \frac{y^2}{36} = 1\Rightarrow \frac{x^2}{7^2} + \frac{y^2}{6^2} = 1

As we can see from the equation, the major axis is along X-axis and the minor axis is along Y-axis.

On comparing the given equation with standard equation of ellipse, which is

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

We get

a=7 and b=6.

So,

c=\sqrt{a^2-b^2}=\sqrt{7^2-6^2}

c=\sqrt{13}

Hence,

Coordinates of the foci:

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

The vertices:

(a,0)\:and\:(-a,0)=(7,0)\:and\:(-7,0)

The length of major axis:

2a=2(7)=14

The length of minor axis:

2b=2(6)=12

The eccentricity :

e=\frac{c}{a}=\frac{\sqrt{13}}{7}

The length of the latus rectum:

\frac{2b^2}{a}=\frac{2(6)^2}{7}=\frac{72}{7}

Question:6 Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse.

\frac{x^2}{100} + \frac{y^2}{400} =1

Answer:

Given

The equation of the ellipse

\frac{x^2}{100} + \frac{y^2}{400} =1\Rightarrow \frac{x^2}{10^2} + \frac{y^2}{20^2} =1

As we can see from the equation, the major axis is along Y-axis and the minor axis is along X-axis.

On comparing the given equation with the standard equation of such ellipse, which is

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

We get

a=20 and b=10.

So,

c=\sqrt{a^2-b^2}=\sqrt{20^2-10^2}

c=\sqrt{300}=10\sqrt{3}

Hence,

Coordinates of the foci:

(0,c)\:and\:(0,-c)=(0,10\sqrt{3})\:and\:(0,-10\sqrt{3})

The vertices:

(0,a)\:and\:(0,-a)=(0,20)\:and\:(0,-20)

The length of the major axis:

2a=2(20)=40

The length of minor axis:

2b=2(10)=20

The eccentricity :

e=\frac{c}{a}=\frac{10\sqrt{3}}{20}=\frac{\sqrt{3}}{2}

The length of the latus rectum:

\frac{2b^2}{a}=\frac{2(10)^2}{20}=\frac{200}{20}=10

Question:7 Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse.

36x^2 + 4y^2 =144

Answer:

Given

The equation of the ellipse

36x^2 + 4y^2 =144

\Rightarrow \frac{36}{144}x^2 + \frac{4}{144}y^2 =1

\Rightarrow \frac{1}{4}x^2 + \frac{1}{36}y^2 =1

\frac{x^2}{2^2} + \frac{y^2}{6^2} = 1

As we can see from the equation, the major axis is along Y-axis and the minor axis is along X-axis.

On comparing the given equation with the standard equation of such ellipse, which is

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

We get

a=6 and b=2.

So,

c=\sqrt{a^2-b^2}=\sqrt{6^2-2^2}

c=\sqrt{32}=4\sqrt{2}

Hence,

Coordinates of the foci:

(0,c)\:and\:(0,-c)=(0,4\sqrt{2})\:and\:(0,-4\sqrt{2})

The vertices:

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

The length of the major axis:

2a=2(6)=12

The length of minor axis:

2b=2(2)=4

The eccentricity :

e=\frac{c}{a}=\frac{4\sqrt{2}}{6}=\frac{2\sqrt{2}}{3}

The length of the latus rectum:

\frac{2b^2}{a}=\frac{2(2)^2}{6}=\frac{8}{6}=\frac{4}{3}

Question:8 Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse.

16x^2 + y^2 = 16

Answer:

Given

The equation of the ellipse

16x^2 + y^2 = 16

\frac{16x^2}{16} + \frac{y^2}{16} = 1

\frac{x^2}{1^2} + \frac{y^2}{4^2} = 1

As we can see from the equation, the major axis is along Y-axis and the minor axis is along X-axis.

On comparing the given equation with the standard equation of such ellipse, which is

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

We get

a=4 and b=1.

So,

c=\sqrt{a^2-b^2}=\sqrt{4^2-1^2}

c=\sqrt{15}

Hence,

Coordinates of the foci:

(0,c)\:and\:(0,-c)=(0,\sqrt{15})\:and\:(0,-\sqrt{15})

The vertices:

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

The length of the major axis:

2a=2(4)=8

The length of minor axis:

2b=2(1)=2

The eccentricity :

e=\frac{c}{a}=\frac{\sqrt{15}}{4}

The length of the latus rectum:

\frac{2b^2}{a}=\frac{2(1)^2}{4}=\frac{2}{4}=\frac{1}{2}

Question:9 Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse.

4x^2 + 9y^2 =36

Answer:

Given

The equation of the ellipse

4x^2 + 9y^2 =36

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

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

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

As we can see from the equation, the major axis is along X-axis and the minor axis is along Y-axis.

On comparing the given equation with the standard equation of an ellipse, which is

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

We get

a=3 and b=2.

So,

c=\sqrt{a^2-b^2}=\sqrt{3^2-2^2}

c=\sqrt{5}

Hence,

Coordinates of the foci:

(c,0)\:and\:(-c,0)=(\sqrt{5},0)\:and\:(-\sqrt{5},0)

The vertices:

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

The length of the major axis:

2a=2(3)=6

The length of minor axis:

2b=2(2)=4

The eccentricity :

e=\frac{c}{a}=\frac{\sqrt{5}}{3}

The length of the latus rectum:

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

Question:10 Find the equation for the ellipse that satisfies the given conditions:

Vertices (± 5, 0), foci (± 4, 0)

Answer:

Given, In an ellipse,

Vertices (± 5, 0), foci (± 4, 0)

Here Vertices and focus of the ellipse are in X-axis so the major axis of this ellipse will be X-axis.

Therefore, the equation of the ellipse will be of the form:

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

Where a and b are the length of the semimajor axis and semiminor axis respectively.

So on comparing standard parameters( vertices and foci) with the given one, we get

a=5 and c=4

Now, As we know the relation,

a^2=b^2+c^2

b^2=a^2-c^2

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

b=\sqrt{5^2-4^2}

b=\sqrt{9}

b=3

Hence, The Equation of the ellipse will be :

\frac{x^2}{5^2}+\frac{y^2}{3^2}=1

Which is

\frac{x^2}{25}+\frac{y^2}{9}=1.

Question:11 Find the equation for the ellipse that satisfies the given conditions:

Vertices (0, ± 13), foci (0, ± 5)

Answer:

Given, In an ellipse,

Vertices (0, ± 13), foci (0, ± 5)

Here Vertices and focus of the ellipse are in Y-axis so the major axis of this ellipse will be Y-axis.

Therefore, the equation of the ellipse will be of the form:

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

Where a and bare the length of the semimajor axis and semiminor axis respectively.

So on comparing standard parameters( vertices and foci) with the given one, we get

a=13 and c=5

Now, As we know the relation,

a^2=b^2+c^2

b^2=a^2-c^2

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

b=\sqrt{13^2-5^2}

b=\sqrt{169-25}

b=\sqrt{144}

b=12

Hence, The Equation of the ellipse will be :

\frac{x^2}{12^2}+\frac{y^2}{13^3}=1

Which is

\frac{x^2}{144}+\frac{y^2}{169}=1.

Question:12 Find the equation for the ellipse that satisfies the given conditions:

Vertices (± 6, 0), foci (± 4, 0)

Answer:

Given, In an ellipse,

Vertices (± 6, 0), foci (± 4, 0)

Here Vertices and focus of the ellipse are in X-axis so the major axis of this ellipse will be X-axis.

Therefore, the equation of the ellipse will be of the form:

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

Where a and bare the length of the semimajor axis and semiminor axis respectively.

So on comparing standard parameters( vertices and foci) with the given one, we get

a=6 and c=4

Now, As we know the relation,

a^2=b^2+c^2

b^2=a^2-c^2

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

b=\sqrt{6^2-4^2}

b=\sqrt{36-16}

b=\sqrt{20}

Hence, The Equation of the ellipse will be :

\frac{x^2}{6^2}+\frac{y^2}{(\sqrt{20})^2}=1

Which is

\frac{x^2}{36}+\frac{y^2}{20}=1.

Question:13 Find the equation for the ellipse that satisfies the given conditions:

Ends of major axis (± 3, 0), ends of minor axis (0, ± 2)

Answer:

Given, In an ellipse,

Ends of the major axis (± 3, 0), ends of minor axis (0, ± 2)

Here, the major axis of this ellipse will be X-axis.

Therefore, the equation of the ellipse will be of the form:

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

Where a and bare the length of the semimajor axis and semiminor axis respectively.

So on comparing standard parameters( ends of the major and minor axis ) with the given one, we get

a=3 and b=2

Hence, The Equation of the ellipse will be :

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

Which is

\frac{x^2}{9}+\frac{y^2}{4}=1.

Question:14 Find the equation for the ellipse that satisfies the given conditions:

Ends of major axis (0, ± \sqrt{5} ), ends of minor axis (± 1, 0)

Answer:

Given, In an ellipse,

Ends of the major axis (0, ±\sqrt{5} ), ends of minor axis (± 1, 0)

Here, the major axis of this ellipse will be Y-axis.

Therefore, the equation of the ellipse will be of the form:

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

Where a and bare the length of the semimajor axis and semiminor axis respectively.

So on comparing standard parameters( ends of the major and minor axis ) with the given one, we get

a=\sqrt{5} and b=1

Hence, The Equation of the ellipse will be :

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

Which is

\frac{x^2}{1}+\frac{y^2}{5}=1.

Question:15 Find the equation for the ellipse that satisfies the given conditions:

Length of major axis 26, foci (± 5, 0)

Answer:

Given, In an ellipse,

Length of major axis 26, foci (± 5, 0)

Here, the focus of the ellipse is in X-axis so the major axis of this ellipse will be X-axis.

Therefore, the equation of the ellipse will be of the form:

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

Where a and bare the length of the semimajor axis and semiminor axis respectively.

So on comparing standard parameters( Length of semimajor axis and foci) with the given one, we get

2a=26\Rightarrow a=13 and c=5

Now, As we know the relation,

a^2=b^2+c^2

b^2=a^2-c^2

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

b=\sqrt{13^2-5^2}

b=\sqrt{144}

b=12

Hence, The Equation of the ellipse will be :

\frac{x^2}{13^2}+\frac{y^2}{12^2}=1

Which is

\frac{x^2}{169}+\frac{y^2}{144}=1.

Question:16 Find the equation for the ellipse that satisfies the given conditions:

Length of minor axis 16, foci (0, ± 6).

Answer:

Given, In an ellipse,

Length of minor axis 16, foci (0, ± 6).

Here, the focus of the ellipse is on the Y-axis so the major axis of this ellipse will be Y-axis.

Therefore, the equation of the ellipse will be of the form:

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

Where a and bare the length of the semimajor axis and semiminor axis respectively.

So on comparing standard parameters( length of semi-minor axis and foci) with the given one, we get

2b=16\Rightarrow b=8 and c=6

Now, As we know the relation,

a^2=b^2+c^2

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

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

a=\sqrt{64+36}

a=\sqrt{100}

a=10

Hence, The Equation of the ellipse will be :

\frac{x^2}{8^2}+\frac{y^2}{10^3}=1

Which is

\frac{x^2}{64}+\frac{y^2}{100}=1.

Question:17 Find the equation for the ellipse that satisfies the given conditions:

Foci (± 3, 0), a = 4

Answer:

Given, In an ellipse,

V Foci (± 3, 0), a = 4

Here foci of the ellipse are in X-axis so the major axis of this ellipse will be X-axis.

Therefore, the equation of the ellipse will be of the form:

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

Where a and bare the length of the semimajor axis and semiminor axis respectively.

So on comparing standard parameters( vertices and foci) with the given one, we get

a=4 and c=3

Now, As we know the relation,

a^2=b^2+c^2

b^2=a^2-c^2

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

b=\sqrt{4^2-3^2}

b=\sqrt{7}

Hence, The Equation of the ellipse will be :

\frac{x^2}{4^2}+\frac{y^2}{(\sqrt{7})^2}=1

Which is

\frac{x^2}{16}+\frac{y^2}{7}=1.

Question:18 Find the equation for the ellipse that satisfies the given conditions:

b = 3, c = 4, centre at the origin; foci on the x axis.

Answer:

Given,In an ellipse,

b = 3, c = 4, centre at the origin; foci on the x axis.

Here foci of the ellipse are in X-axis so the major axis of this ellipse will be X-axis.

Therefore, the equation of the ellipse will be of the form:

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

Where a and bare the length of the semimajor axis and semiminor axis respectively.

Also Given,

b=3 and c=4

Now, As we know the relation,

a^2=b^2+c^2

a^2=3^2+4^2

a^2=25

a=5

Hence, The Equation of the ellipse will be :

\frac{x^2}{5^2}+\frac{y^2}{3^2}=1

Which is

\frac{x^2}{25}+\frac{y^2}{9}=1.

Question:19 Find the equation for the ellipse that satisfies the given conditions:

Centre at (0,0), major axis on the y-axis and passes through the points (3, 2) and (1,6).

Answer:

Given,in an ellipse

Centre at (0,0), major axis on the y-axis and passes through the points (3, 2) and (1,6).

Since, The major axis of this ellipse is on the Y-axis, the equation of the ellipse will be of the form:

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

Where a and b are the length of the semimajor axis and semiminor axis respectively.

Now since the ellipse passes through points,(3, 2)

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

{9a^2+4b^2}={a^2b^2}

since the ellipse also passes through points,(1, 6).

\frac{1^2}{b^2}+\frac{6^2}{a^2}=1

a^2+36b^2=a^2b^2

On solving these two equation we get

a^2=40 and b^2=10

Thus, The equation of the ellipse will be

\frac{x^2}{10}+\frac{y^2}{40}=1.

Question:20 Find the equation for the ellipse that satisfies the given conditions:

Major axis on the x-axis and passes through the points (4,3) and (6,2).

Answer:

Given, in an ellipse

Major axis on the x-axis and passes through the points (4,3) and (6,2).

Since The major axis of this ellipse is on the X-axis, the equation of the ellipse will be of the form:

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

Where a and bare the length of the semimajor axis and semiminor axis respectively.

Now since the ellipse passes through the point,(4,3)

\frac{4^2}{a^2}+\frac{3^2}{b^2}=1

{16b^2+9a^2}={a^2b^2}

since the ellipse also passes through the point (6,2).

\frac{6^2}{a^2}+\frac{2^2}{b^2}=1

4a^2+36b^2=a^2b^2

On solving this two equation we get

a^2=52 and b^2=13

Thus, The equation of the ellipse will be

\frac{x^2}{52}+\frac{y^2}{13}=1

More About NCERT Solutions for Class 11 Maths Chapter 11 Exercise 11.3:-

Class 11 Maths chapter 11 exercise 11.3 consists of questions related to finding the coordinates of the foci of an ellipse, the vertices of an ellipse, the length of the major axis and minor axis of an ellipse, the eccentricity and the length of the latus rectum of an ellipse given the equation of an ellipse. Also, you will learn to write the equation of the ellipse when the other parameters of the ellipse are given.

Also Read| Conic Section Class 11 Notes

Benefits of NCERT Solutions for Class 11 Maths Chapter 11 Exercise 11.3:-

  • There are five solved examples given before the exercise 11.3 Class 11 Maths that you can try to solve in order to get familiar with ellipse concepts.
  • NCERT solutions for Class 11 Maths chapter 11 exercise 11.3 are here to help you when you face difficulty while solving NCERT problems.
  • You can use Class 11 Maths chapter 11 exercise 11.3 solutions as a reference while solving them.

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 definition of ellipse ?

The set of all points in a plane, the sum of whose distances from two fixed points in the plane is a constant is called an ellipse.

2. What is the centre of the ellipse ?

The centre of the ellipse is the mid-point of the line segment joining the foci.

3. What is the major axis of the ellipse ?

The major axis of an ellipse is a line segment passing through the foci of the ellipse.

4. What is the minor axis of the ellipse ?

The minor axis of an ellipse is a line segment passing through the centre and perpendicular to the major axis.

5. What is the latus rectum of an ellipse ?

The Latus rectum of an ellipse is a line segment passing through the foci and perpendicular to the major axis whose endpoints lie on the ellipse.

6. What is the weightage on conic section in the CBSE Class 11 Maths ?

The whole coordinate geometry unit has 10 marks weightage in the CBSE Class 11 Maths final exam.

Also Read

Articles

Upcoming School Exams

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

Mizoram Board of School Education 12th Examination

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

Mizoram Board 10th Examination

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

Punjab Board of Secondary Education 10th Examination

Application Correction Date:08 October,2024 - 27 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
Apply

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

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
Apply

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

TOEFL ® Registrations 2024
Apply

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

Pearson | PTE
Apply

Register now for PTE & Unlock 20% OFF : Use promo code: 'C360SPL20'. Valid till 30th NOV'24! Trusted by 3,500+ universities globally

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
MP: Private schools to pay back illegally collected tuition fees worth Rs 31 crores
Tamil Nadu teacher fatally stabbed in school, violence erupts; assailant detained
Bihar govt puts hold on new teachers transfer policy
Youth stabs teacher on school premises; now detained, says police
Air Pollution: Schools in Rajasthan's Khairthal-Tijara go online for classes 1 to 5
Govt informs SC about NCERT guidelines for development of e-content for children with disabilities
Government school principal in Andhra chops off girl students' hair for coming late
Back to top