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

The solutions for the questions present in the third exercise of Chapter 11, Class 11, maths are given here. These PDFs can be downloaded from the links given below. Exercise 11.3 of NCERT Solutions for Class 11 Maths Chapter 11- Conic Sections is based on the following topics:

1. Ellipse
   1. Relationship between 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 BYJU’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.

Download PDF of NCERT Solutions for Class 11 Maths Chapter 11- Conic Sections Exercise 11.3

Solutions for Class 11 Maths Chapter 11 – 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 x²/36 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 with x²/a² + y²/b² = 1, we get

a = 6 and b = 4.

c = √(a² – b²)

= √(36-16)

= √20

= 2√5

Then,

The coordinates of the foci are (2√5, 0) and (-2√5, 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, e = c/a = 2√5/6 = √5/3

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

2. x²/4 + y²/25 = 1

Solution:

Given:

The equation is x²/4 + y²/25 = 1

Here, the denominator of y²/25 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 = √(a² – b²)

= √(25-4)

= √21

Then,

The coordinates of the foci are (0, √21) and (0, -√21).

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/a = √21/5

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

3.  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 x²/16 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 with x²/a² + y²/b² = 1, we get

a = 4 and b = 3.

c = √(a² – b²)

= √(16-9)

= √7

Then,

The coordinates of the foci are (√7, 0) and (-√7, 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/a = √7/4

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

4.  x²/25 + y²/100 = 1

Solution:

Given:

The equation is x²/25 + y²/100 = 1

Here, the denominator of y²/100 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 with x²/b² + y²/a² = 1, we get

b = 5 and a =10.

c = √(a² – b²)

= √(100-25)

= √75

= 5√3

Then,

The coordinates of the foci are (0, 5√3) and (0, -5√3).

The coordinates of the vertices are (0, √10) and (0, -√10)

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

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

Eccentricity, e = c/a = 5√3/10 = √3/2

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

5. x²/49 + y²/36 = 1

Solution:

Given:

The equation is x²/49 + y²/36 = 1

Here, the denominator of x²/49 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 with x²/a² + y²/b² = 1, we get

b = 6 and a =7

c = √(a² – b²)

= √(49-36)

= √13

Then,

The coordinates of the foci are (√13, 0) and (-√3, 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/a = √13/7

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

6. x²/100 + y²/400 = 1

Solution:

Given:

The equation is x²/100 + y²/400 = 1

Here, the denominator of y²/400 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 with x²/b² + y²/a² = 1, we get

b = 10 and a =20.

c = √(a² – b²)

= √(400-100)

= √300

= 10√3

Then,

The coordinates of the foci are (0, 10√3) and (0, -10√3).

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/a = 10√3/20 = √3/2

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

7. 36x² + 4y² = 144

Solution:

Given:

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

Here, the denominator of y²/6² 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 with x²/b² + y²/a² = 1, we get

b = 2 and a = 6.

c = √(a² – b²)

= √(36-4)

= √32

= 4√2

Then,

The coordinates of the foci are (0, 4√2) and (0, -4√2).

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/a = 4√2/6 = 2√2/3

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

8. 16x² + y² = 16

Solution:

Given:

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

Here, the denominator of y²/4² 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 with x²/b² + y²/a² = 1, we get

b =1 and a =4.

c = √(a² – b²)

= √(16-1)

= √15

Then,

The coordinates of the foci are (0, √15) and (0, -√15).

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/a = √15/4

Length of latus rectum = 2b²/a = (2×1²)/4 = 2/4 = ½

9. 4x² + 9y² = 36

Solution:

Given:

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

Here, the denominator of x²/3² 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 with x²/a² + y²/b² = 1, we get

a =3 and b =2.

c = √(a² – b²)

= √(9-4)

= √5

Then,

The coordinates of the foci are (√5, 0) and (-√5, 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/a = √5/3

Length of latus rectum = 2b²/a = (2×2²)/3 = (2×4)/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 (± 5, 0), foci (± 4, 0)

Solution:

Given:

Vertices (± 5, 0) and foci (± 4, 0)

Here, the vertices are on the x-axis.

So, the equation of the ellipse will be of the form x²/a² + y²/b² = 1, where ‘a’ is the semi-major axis.

Then, a = 5 and c = 4.

It is known that a² = b² + c².

So, 5² = b² + 4²

25 = b² + 16

b² = 25 – 16

b = √9

= 3

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

11. Vertices (0, ± 13), foci (0, ± 5)

Solution:

Given:

Vertices (0, ± 13) and foci (0, ± 5)

Here, the vertices are on the y-axis.

So, the equation of the ellipse will be of the form x²/b² + y²/a² = 1, where ‘a’ is the semi-major axis.

Then, a =13 and c = 5.

It is known that a² = b² + c².

13² = b²+5²

169 = b² + 15

b² = 169 – 125

b = √144

= 12

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

12. Vertices (± 6, 0), foci (± 4, 0)

Solution:

Given:

Vertices (± 6, 0) and foci (± 4, 0)

Here, the vertices are on the x-axis.

So, the equation of the ellipse will be of the form x²/a² + y²/b² = 1, where ‘a’ is the semi-major axis.

Then, a = 6 and c = 4.

It is known that a² = b² + c².

6² = b²+4²

36 = b² + 16

b² = 36 – 16

b = √20

∴ The equation of the ellipse is x²/6² + y²/(√20)² = 1 or x²/36 + y²/20 = 1

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

Solution:

Given:

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

Here, 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 ‘a’ is the semi-major axis.

Then, a = 3 and b = 2.

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

14. Ends of major axis (0, ±√5), ends of minor axis (±1, 0)

Solution:

Given:

Ends of major axis (0, ±√5) and ends of minor axis (±1, 0)

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

So, the equation of the ellipse will be of the form x²/b² + y²/a² = 1, where ‘a’ is the semi-major axis.

Then, a = √5 and b = 1.

∴ The equation for the ellipse x²/1² + y²/(√5)² = 1 or x²/1 + y²/5 = 1

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

Solution:

Given:

Length of major axis is 26 and foci (±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 form x²/a² + y²/b² = 1, where ‘a’ is the semi-major axis.

Then, 2a = 26

a = 13 and c = 5.

It is known that a² = b² + c².

13² = b²+5²

169 = b² + 25

b² = 169 – 25

b = √144

= 12

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

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

Solution:

Given:

Length of minor axis is 16 and foci (0, ±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 form x²/b² + y²/a² = 1, where ‘a’ is the semi-major axis.

Then, 2b =16

b = 8 and c = 6.

It is known that a² = b² + c².

a² = 8² + 6²

= 64 + 36

=100

a = √100

= 10

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

17. Foci (±3, 0), a = 4

Solution:

Given:

Foci (±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 ‘a’ is the semi-major axis.

Then, c = 3 and a = 4.

It is known that a² = b² + c².

a² = 8² + 6²

= 64 + 36

= 100

16 = b² + 9

b² = 16 – 9

= 7

∴ 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 form x²/a² + y²/b² = 1, where ‘a’ is the semi-major axis.

Then, b = 3 and c = 4.

It is known that a² = b² + c².

a² = 3² + 4²

= 9 + 16

=25

a = √25

= 5

∴ 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 ‘a’ 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/a²…. (1)

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

1¹/b² + 6²/a² = 1

1/b² + 36/a² = 1 …. (2)

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

b² = 10 and a² = 40.

∴ The equation of the ellipse is x²/10 + y²/40 = 1 or 4x² + y ² = 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 = 1…. (1) [Where ‘a’ 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 …. (2)

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

36/a² + 4/b² = 1 …. (3)

From equation (2)

16/a² = 1 – 9/b²

1/a² = (1/16 (1 – 9/b²)) …. (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 – 9/b²)] + 4/b² = 1

36/16 (1 – 9/b²) + 4/b² = 1

9/4 (1 – 9/b²) + 4/b² = 1

9/4 – 81/4b² + 4/b² = 1

-81/4b² + 4/b² = 1 – 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 – 9/b²)

= 1/16(1 – 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 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

Miscellaneous Exercise On Chapter 11 Solutions 8 Questions