# NCERT Solutions for Class 11 Maths Chapter 11 - Conic Sections Exercise 11.3

*According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 10.

The NCERT Solutions for the questions in the third exercise of Chapter 11, Class 11 Maths are given here. Chapter 11 Conic Sections of Class 11 Maths is included in the CBSE Syllabus 2023-24. The PDF version can be downloaded from the links given below. Exercise 11.3 of NCERT Solutions for Class 11 Maths Chapter 11 â€“ Conic Sections is based on the following topics:

1. Ellipse
1. Relationship between the semi-major axis, semi-minor axis and the distance of the focus from the centre of the ellipse
2. Special cases of an ellipse
3. Eccentricity
4. Standard equations of an ellipse
5. Latus rectum

These solutions are prepared by subject matter experts at BYJUâ€™S, describing the complete method of solving problems. By understanding the concepts used in NCERT Solutions for Class 11 Maths, students will be able to clear all their doubts related to topics present in Class 11 Maths and ace the board exam.

## NCERT Solutions for Class 11 Maths Chapter 11 â€“ Conic Sections Exercise 11.3

### Solutions for Class 11 Maths Chapter 11 â€“ Exercise 11.3

In each of the Exercises 1 to 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.

1. x2/36 + y2/16 = 1

Solution:

Given:

The equation is x2/36 + y2/16 = 1

Here, the denominator of x2/36Â is greater than the denominator of y2/16.

So, the major axis is along the x-axis, while the minor axis is along the y-axis.

On comparing the given equation withÂ x2/a2 + y2/b2 = 1, we get

a = 6 and b = 4.

c = âˆš(a2Â â€“ b2)

= âˆš(36-16)

= âˆš20

= 2âˆš5

Then,

The coordinates of the foci are (2âˆš5, 0) and (-2âˆš5, 0).

The coordinates of the vertices are (6, 0) and (-6, 0)

Length of major axis = 2a = 2 (6) = 12

Length of minor axis = 2b = 2 (4) = 8

Eccentricity, eÂ = c/aÂ = 2âˆš5/6 = âˆš5/3

Length of latus rectum = 2b2/a = (2Ã—16)/6 = 16/3

2. x2/4 + y2/25 = 1

Solution:

Given:

The equation is x2/4 + y2/25 = 1

Here, the denominator of y2/25Â is greater than the denominator of x2/4.

So, the major axis is along the x-axis, while the minor axis is along the y-axis.

On comparing the given equation with x2/a2 + y2/b2 = 1, we get

a = 5 and b = 2.

c = âˆš(a2Â â€“ b2)

= âˆš(25-4)

= âˆš21

Then,

The coordinates of the foci are (0, âˆš21) and (0, -âˆš21).

The coordinates of the vertices are (0, 5) and (0, -5)

Length of major axis = 2a = 2 (5) = 10

Length of minor axis = 2b = 2 (2) = 4

Eccentricity, e = c/aÂ = âˆš21/5

Length of latus rectum = 2b2/a = (2Ã—22)/5 = (2Ã—4)/5 = 8/5

3. Â x2/16 + y2/9 = 1

Solution:

Given:

The equation is x2/16 + y2/9 = 1 or x2/42 + y2/32 = 1

Here, the denominator of x2/16Â is greater than the denominator of y2/9.

So, the major axis is along the x-axis, while the minor axis is along the y-axis.

On comparing the given equation withÂ x2/a2 + y2/b2 = 1, we get

a = 4 and b = 3.

c = âˆš(a2Â â€“ b2)

= âˆš(16-9)

= âˆš7

Then,

The coordinates of the foci are (âˆš7, 0) and (-âˆš7, 0).

The coordinates of the vertices are (4, 0) and (-4, 0)

Length of major axis = 2a = 2 (4) = 8

Length of minor axis = 2b = 2 (3) = 6

Eccentricity, e = c/aÂ = âˆš7/4

Length of latus rectum = 2b2/a = (2Ã—32)/4 = (2Ã—9)/4 = 18/4 = 9/2

4. Â x2/25 + y2/100 = 1

Solution:

Given:

The equation is x2/25 + y2/100 = 1

Here, the denominator of y2/100Â is greater than the denominator of x2/25.

So, the major axis is along the y-axis, while the minor axis is along the x-axis.

On comparing the given equation withÂ x2/b2 + y2/a2 = 1, we get

b = 5 and a =10.

c = âˆš(a2Â â€“ b2)

= âˆš(100-25)

= âˆš75

= 5âˆš3

Then,

The coordinates of the foci are (0, 5âˆš3) and (0, -5âˆš3).

The coordinates of the vertices are (0, âˆš10) and (0, -âˆš10)

Length of major axis = 2a = 2 (10) = 20

Length of minor axis = 2b = 2 (5) = 10

Eccentricity, e = c/aÂ = 5âˆš3/10 = âˆš3/2

Length of latus rectum = 2b2/a = (2Ã—52)/10 = (2Ã—25)/10 = 5

5. x2/49 + y2/36 = 1

Solution:

Given:

The equation is x2/49 + y2/36 = 1

Here, the denominator of x2/49Â is greater than the denominator of y2/36.

So, the major axis is along the x-axis, while the minor axis is along the y-axis.

On comparing the given equation withÂ x2/a2 + y2/b2 = 1, we get

b = 6 and a =7

c = âˆš(a2Â â€“ b2)

= âˆš(49-36)

= âˆš13

Then,

The coordinates of the foci are (âˆš13, 0) and (-âˆš3, 0).

The coordinates of the vertices are (7, 0) and (-7, 0)

Length of major axis = 2a = 2 (7) = 14

Length of minor axis = 2b = 2 (6) = 12

Eccentricity, e = c/aÂ = âˆš13/7

Length of latus rectum = 2b2/a = (2Ã—62)/7 = (2Ã—36)/7 = 72/7

6. x2/100 + y2/400 = 1

Solution:

Given:

The equation is x2/100 + y2/400 = 1

Here, the denominator of y2/400Â is greater than the denominator of x2/100.

So, the major axis is along the y-axis, while the minor axis is along the x-axis.

On comparing the given equation withÂ x2/b2 + y2/a2 = 1, we get

b = 10 and a =20.

c = âˆš(a2Â â€“ b2)

= âˆš(400-100)

= âˆš300

= 10âˆš3

Then,

The coordinates of the foci are (0, 10âˆš3) and (0, -10âˆš3).

The coordinates of the vertices are (0, 20) and (0, -20)

Length of major axis = 2a = 2 (20) = 40

Length of minor axis = 2b = 2 (10) = 20

Eccentricity, e = c/aÂ = 10âˆš3/20 = âˆš3/2

Length of latus rectum = 2b2/a = (2Ã—102)/20 = (2Ã—100)/20 = 10

7. 36x2Â + 4y2Â = 144

Solution:

Given:

The equation is 36x2Â + 4y2Â = 144 or x2/4 + y2/36 = 1 or x2/22 + y2/62 = 1

Here, the denominator of y2/62Â is greater than the denominator of x2/22.

So, the major axis is along the y-axis, while the minor axis is along the x-axis.

On comparing the given equation withÂ x2/b2 + y2/a2 = 1, we get

b = 2 and a = 6.

c = âˆš(a2Â â€“ b2)

= âˆš(36-4)

= âˆš32

= 4âˆš2

Then,

The coordinates of the foci are (0, 4âˆš2) and (0, -4âˆš2).

The coordinates of the vertices are (0, 6) and (0, -6)

Length of major axis = 2a = 2 (6) = 12

Length of minor axis = 2b = 2 (2) = 4

Eccentricity, e = c/aÂ = 4âˆš2/6 = 2âˆš2/3

Length of latus rectum = 2b2/a = (2Ã—22)/6 = (2Ã—4)/6 = 4/3

8. 16x2Â + y2Â = 16

Solution:

Given:

The equation is 16x2Â + y2Â = 16 or x2/1 + y2/16 = 1 or x2/12 + y2/42 = 1

Here, the denominator of y2/42Â is greater than the denominator of x2/12.

So, the major axis is along the y-axis, while the minor axis is along the x-axis.

On comparing the given equation withÂ x2/b2 + y2/a2 = 1, we get

b =1 and a =4.

c = âˆš(a2Â â€“ b2)

= âˆš(16-1)

= âˆš15

Then,

The coordinates of the foci are (0, âˆš15) and (0, -âˆš15).

The coordinates of the vertices are (0, 4) and (0, -4)

Length of major axis = 2a = 2 (4) = 8

Length of minor axis = 2b = 2 (1) = 2

Eccentricity, e = c/aÂ = âˆš15/4

Length of latus rectum = 2b2/a = (2Ã—12)/4 = 2/4 = Â½

9. 4x2Â + 9y2Â = 36

Solution:

Given:

The equation is 4x2Â + 9y2Â = 36 or x2/9 + y2/4 = 1 or x2/32 + y2/22 = 1

Here, the denominator of x2/32Â is greater than the denominator of y2/22.

So, the major axis is along the x-axis, while the minor axis is along the y-axis.

On comparing the given equation withÂ x2/a2 + y2/b2 = 1, we get

a = 3 and b = 2.

c = âˆš(a2Â â€“ b2)

= âˆš(9-4)

= âˆš5

Then,

The coordinates of the foci are (âˆš5, 0) and (-âˆš5, 0).

The coordinates of the vertices are (3, 0) and (-3, 0)

Length of major axis = 2a = 2 (3) = 6

Length of minor axis = 2b = 2 (2) = 4

Eccentricity, e = c/aÂ = âˆš5/3

Length of latus rectum = 2b2/a = (2Ã—22)/3 = (2Ã—4)/3 = 8/3

In each of the following Exercises 10 to 20, find the equation for the ellipse that satisfies the given conditions:

10. Vertices (Â±Â 5, 0), foci (Â±Â 4, 0)

Solution:

Given:

Vertices (Â±Â 5, 0) and foci (Â±Â 4, 0)

Here, the vertices are on the x-axis.

So, the equation of the ellipse will be of the formÂ x2/a2 + y2/b2 = 1, where â€˜aâ€™ is the semi-major axis.

Then, a = 5 and c = 4.

It is known that a2Â = b2 + c2.

So, 52Â = b2 + 42

25 = b2Â + 16

b2Â = 25 â€“ 16

b =Â âˆš9

= 3

âˆ´ The equation of the ellipse is x2/52 + y2/32 = 1 or x2/25 + y2/9Â = 1

11. Vertices (0,Â Â±Â 13), foci (0,Â Â± 5)

Solution:

Given:

Vertices (0,Â Â±Â 13) and foci (0,Â Â± 5)

Here, the vertices are on the y-axis.

So, the equation of the ellipse will be of the formÂ x2/b2 + y2/a2 = 1, where â€˜aâ€™ is the semi-major axis.

Then, a =13 and c = 5.

It is known that a2Â = b2 + c2.

132Â = b2+52

169 = b2Â + 15

b2Â = 169 â€“ 125

b =Â âˆš144

= 12

âˆ´ The equation of the ellipse is x2/122 + y2/132Â = 1 or x2/144 + y2/169Â = 1

12. Vertices (Â±Â 6, 0), foci (Â± 4, 0)

Solution:

Given:

Vertices (Â±Â 6, 0) and foci (Â± 4, 0)

Here, the vertices are on the x-axis.

So, the equation of the ellipse will be of the formÂ x2/a2 + y2/b2 = 1, where â€˜aâ€™ is the semi-major axis.

Then, a = 6 and c = 4.

It is known that a2Â = b2 + c2.

62Â = b2+42

36 = b2Â + 16

b2Â = 36 â€“ 16

b =Â âˆš20

âˆ´ The equation of the ellipse is x2/62 + y2/(âˆš20)2Â = 1 or x2/36 + y2/20Â = 1

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

Solution:

Given:

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

Here, the major axis is along the x-axis.

So, the equation of the ellipse will be of the formÂ x2/a2 + y2/b2 = 1, where â€˜aâ€™ is the semi-major axis.

Then, a = 3 and b = 2.

âˆ´ The equation for the ellipse x2/32 + y2/22Â = 1 or x2/9 + y2/4Â = 1

14. Ends of major axis (0,Â Â±âˆš5), ends of minor axis (Â±1, 0)

Solution:

Given:

Ends of major axis (0,Â Â±âˆš5) and ends of minor axis (Â±1, 0)

Here, the major axis is along the y-axis.

So, the equation of the ellipse will be of the formÂ x2/b2 + y2/a2 = 1, where â€˜aâ€™ is the semi-major axis.

Then, a =Â âˆš5 and b = 1.

âˆ´ The equation for the ellipse x2/12 + y2/(âˆš5)2Â = 1 or x2/1 + y2/5Â = 1

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

Solution:

Given:

Length of the major axis is 26 and the foci (Â±5, 0)

Since the foci are on the x-axis, the major axis is along the x-axis.

So, the equation of the ellipse will be of the formÂ x2/a2 + y2/b2 = 1, where â€˜aâ€™ is the semi-major axis.

Then, 2a = 26

a = 13 and c = 5.

It is known that a2Â = b2 + c2.

132Â = b2+52

169 = b2Â + 25

b2Â = 169 â€“ 25

b =Â âˆš144

= 12

âˆ´ The equation of the ellipse is x2/132 + y2/122 = 1Â or x2/169 + y2/144Â = 1

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

Solution:

Given:

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

Since the foci are on the y-axis, the major axis is along the y-axis.

So, the equation of the ellipse will be of the formÂ x2/b2 + y2/a2 = 1, where â€˜aâ€™ is the semi-major axis.

Then, 2b =16

b = 8 and c = 6.

It is known that a2Â = b2 + c2.

a2Â = 82 + 62

=Â 64 + 36

=100

a =Â âˆš100

= 10

âˆ´ The equation of the ellipse is x2/82 + y2/102Â =1 or x2/64 + y2/100Â = 1

17. Foci (Â±3, 0), a = 4

Solution:

Given:

Foci (Â±3, 0) and a = 4

Since the foci are on the x-axis, the major axis is along the x-axis.

So, the equation of the ellipse will be of the form x2/a2 + y2/b2 = 1, where â€˜aâ€™ is the semi-major axis.

Then, c = 3 and a = 4.

It is known that a2Â = b2 + c2.

a2Â = 82 + 62

= 64 + 36

= 100

16 = b2Â + 9

b2Â = 16 â€“ 9

= 7

âˆ´ The equation of the ellipse is x2/16 + y2/7 = 1

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

Solution:

Given:

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

Since the foci are on the x-axis, the major axis is along the x-axis.

So, the equation of the ellipse will be of the formÂ x2/a2 + y2/b2 = 1, where â€˜aâ€™ is the semi-major axis.

Then, b = 3 and c = 4.

It is known that a2Â = b2 + c2.

a2Â = 32 + 42

= 9 + 16

=25

a = âˆš25

= 5

âˆ´ The equation of the ellipse is x2/52 + y2/32 or x2/25 + y2/9 = 1

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

Solution:

Given:

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

Since the centre is at (0, 0) and the major axis is on the y- axis, the equation of the ellipse will be of the form x2/b2 + y2/a2 = 1, where â€˜aâ€™ is the semi-major axis.

The ellipse passes through points (3, 2) and (1, 6).

So, by putting the values x = 3 and y = 2, we get,

32/b2 + 22/a2 = 1

9/b2 + 4/a2â€¦. (1)

And by putting the values x = 1 and y = 6, we get,

11/b2 + 62/a2 = 1

1/b2 + 36/a2 = 1 â€¦. (2)

On solving equations (1) and (2), we get

b2Â = 10 and a2Â = 40.

âˆ´ The equation of the ellipse is x2/10 + y2/40Â = 1 or 4x2Â + yÂ 2Â = 40

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

Solution:

Given:

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

Since the major axis is on the x-axis, the equation of the ellipse will be the form

x2/a2 + y2/b2 = 1â€¦. (1) [Where â€˜aâ€™ is the semi-major axis.]

The ellipse passes through points (4, 3) and (6, 2).

So by putting the values x = 4 and y = 3 in equation (1), we get,

16/a2 + 9/b2 = 1 â€¦. (2)

Putting x = 6 and y = 2 in equation (1), we get,

36/a2 + 4/b2 = 1 â€¦. (3)

From equation (2)

16/a2 = 1 â€“ 9/b2

1/a2 = (1/16 (1 â€“ 9/b2)) â€¦. (4)

Substituting the value of 1/a2 in equation (3), we get,

36/a2 + 4/b2 = 1

36(1/a2) + 4/b2 = 1

36[1/16 (1 â€“ 9/b2)] + 4/b2 = 1

36/16 (1 â€“ 9/b2) + 4/b2 = 1

9/4 (1 â€“ 9/b2) + 4/b2 = 1

9/4 â€“ 81/4b2 + 4/b2 = 1

-81/4b2 + 4/b2 = 1 â€“ 9/4

(-81+16)/4b2 = (4-9)/4

-65/4b2 = -5/4

-5/4(13/b2) = -5/4

13/b2 = 1

1/b2 = 1/13

b2 = 13

Now substitute the value of b2 in equation (4) we get,

1/a2 = 1/16(1 â€“ 9/b2)

= 1/16(1 â€“ 9/13)

= 1/16((13-9)/13)

= 1/16(4/13)

= 1/52

a2 = 52

Equation of ellipse is x2/a2 + y2/b2 = 1

By substituting the values of a2 and b2 in the above equation, we get,

x2/52 + y2/13 = 1

### Access Other Exercise Solutions of Class 11 Maths Chapter 11- Conic Sections

Exercise 11.1 Solutions 15 Questions

Exercise 11.2 Solutions 12 Questions

Exercise 11.4 Solutions 15 Questions

Also explore â€“ NCERT Class 11 Solutions