# Solutions for Class 11 Maths Chapter 11 – Exercise 11.3

## Solutions for Class 11 Maths Chapter 11

In each of 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 = b+ c2.

So, 52 = b+ 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 = b+ 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 = b+ 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 = b+ 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 = b+ c2.

a2 = 8+ 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 = b+ c2.

a2 = 8+ 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 = b+ c2.

a2 = 3+ 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 the above equation we get,

x2/52 + y2/13 = 1

Solutions for Class 11 Maths Chapter 11 – Exercise 11.3
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