# Class 11 NCERT Solutions- Chapter 11 Conic Section – Exercise 11.3

Last Updated : 09 Mar, 2021

### Question 1. = 1

Solution:

Since denominator of x2/36 is larger than the denominator of y2/16,

the major axis is along the x-axis.

On comparing the given equation with  = 1, we get

a2 = 36 and b2 = 16

â‡’ a = Â±6 and b = Â±4

The Foci:

Foci = (c, 0) and (-c, 0) when (a2 > b2)

c = âˆš(a2 – b2)          -(when a2 > b2)

c = âˆš(36 – 16)

c = âˆš20 = 2âˆš5

â‡’ (2âˆš5, 0) and (-2âˆš5, 0)

Vertices:

Vertices = (a, 0) and (-a, 0) when (a2 > b2)

â‡’ (6, 0) and (-6, 0)

Length of major axis:

Length of major axis = 2a (when a2 > b2)

= 2 Ã— 6

â‡’ Length of major axis = 12

Length of minor axis:

Length of minor axis = 2b (when a2 > b2)

= 2 Ã— 4

â‡’ Length of minor axis = 8

Eccentricity

Eccentricity = c/a (when a2 > b2)

= 2âˆš5/6

= âˆš5/3

Length of the latus rectum:

Length of the latus rectum = 2b2/a (when a2 > b2)

= 2Ã—16/6

= 16/3

### Question 2.  = 1

Solution:

Since denominator of y2/25 is larger than the denominator of x2/4,

the major axis is along the y-axis.

Comparing the given equation with  = 1, we get

a2 = 4 and b2 = 25

â‡’ a = Â±2 and b = Â±5

The Foci:

Foci = (0, c) and (0, -c) when (a2 < b2)

c = âˆš(b2 – a2)       -(when a2 < b2)

c = âˆš(25 – 4)

c = âˆš21

â‡’ (0, âˆš21) and (0, -âˆš21)

Vertices:

Vertices = (0, b) and (0, -b) when (a2 < b2)

â‡’ (0, 5) and (0, -5)

Length of major axis:

Length of major axis = 2b (when a2 < b2)

= 2 Ã— 5

â‡’ Length of major axis = 10

Length of minor axis:

Length of minor axis = 2a (when a2 < b2)

=2 Ã— 2

â‡’ Length of minor axis = 4

Eccentricity:

Eccentricity = c/b (when a2 < b2)

= âˆš21/5

Length of the latus rectum:

Length of the latus rectum = 2a2/b (when a2 < b2)

= 2Ã—4/5

= 8/5

### Question 3.  = 1

Solution:

Since denominator of x2/16 is larger than the denominator of y2/9,

the major axis is along the x-axis.

Comparing the given equation with  = 1, we get

a2 = 16 and b2 = 9

â‡’ a = Â±4 and b = Â±3

The Foci:

Foci = (c, 0) and (-c, 0) when (a2 > b2)

c = âˆš(a2 – b2)       -(when a2 > b2)

c = âˆš(16 – 9)

c = âˆš7

â‡’ (âˆš7, 0) and (-âˆš7, 0).

Vertices:

Vertices = (a, 0) and (-a, 0) when (a2 > b2)

â‡’ (4,0) and (-4,0)

Length of major axis:

Length of major axis = 2a (when a2 > b2)

= 2 Ã— 4

â‡’ Length of major axis = 8

Length of minor axis:

Length of minor axis = 2b (when a2 > b2)

= 2 Ã— 3

â‡’ Length of minor axis = 6

Eccentricity:

Eccentricity = c/a (when a2>b2)

= âˆš7/4

Length of the latus rectum:

Length of the latus rectum = 2b2/a (when a2 > b2)

= 2 Ã— 9/4

= 9/2

### Question 4.  = 1

Solution:

Since denominator of y2/100 is larger than the denominator of x2/25,

the major axis is along the y-axis.

Comparing the given equation with  = 1, we get

a2 = 25 and b2 = 100

â‡’ a = Â±5 and b = Â±10

The Foci:

Foci = (0, c) and (0, -c) when (a2 < b2)

c = âˆš(b2 – a2)       -(when a2 < b2)

c = âˆš(100 – 25)

c = âˆš75

c = 5âˆš3

â‡’ (0, 5âˆš3) and (0, -5âˆš3)

Vertices:

Vertices = (0, b) and (0, -b) when (a2 < b2)

â‡’ (0, 10) and (0, -10)

Length of major axis:

Length of major axis = 2b (when a2 < b2)

= 2 Ã— 10

â‡’ Length of major axis = 20

Length of minor axis:

Length of minor axis = 2a (when a2 < b2)

= 2 Ã— 5

â‡’ Length of minor axis = 10

Eccentricity:

Eccentricity = c/b (when a2 < b2)

= 5âˆš3/10

= âˆš3/2

Length of the latus rectum:

Length of the latus rectum = 2a2/b (when a2 < b2)

= 2 Ã— 25/10

= 5

### Question 5.  = 1

Solution:

Since denominator of x2/49 is larger than the denominator of y2/36,

the major axis is along the x-axis.

Comparing the given equation with  = 1, we get

a2 = 49 and b2 = 36

â‡’ a = Â±7 and b = Â±6

The Foci:

Foci = (c, 0) and (-c, 0) when (a2 > b2)

c = âˆš(a2 – b2)       -(when a2 > b2)

c = âˆš(49 – 36)

c = âˆš13

â‡’ (âˆš13, 0) and (-âˆš13, 0).

Vertices:

Vertices = (a, 0) and (-a, 0) when (a2 > b2)

â‡’ (7, 0) and (-7, 0)

Length of major axis:

Length of major axis = 2a (when a2 > b2)

= 2 Ã— 7

â‡’ Length of major axis = 14

Length of minor axis:

Length of minor axis = 2b (when a2 > b2)

= 2 Ã— 6

â‡’ Length of minor axis = 12

Eccentricity:

Eccentricity = c/a (when a2 > b2)

= âˆš13/7

Length of the latus rectum:

Length of the latus rectum = 2b2/a (when a2 > b2)

= 2 Ã— 36/7

= 72/7

### Question 6. = 1

Solution:

Since denominator of y2/400 is larger than the denominator of x2/100,

the major axis is along the y-axis.

Comparing the given equation with  = 1, we get

a2 = 100 and b2 = 400

â‡’ a = Â±10 and b = Â±20

The Foci:

Foci = (0, c) and (0, -c) when (a2 < b2)

c = âˆš(b2 – a2)       -(when a2 < b2)

c = âˆš(400 – 100)

c = âˆš300

c = 10âˆš3

â‡’ (0, 10âˆš3) and (0, -10âˆš3)

Vertices:

Vertices = (0, b) and (0, -b) when (a2 < b2)

â‡’ (0, 20) and (0, -20)

Length of major axis:

Length of major axis = 2b (when a2 < b2)

= 2 Ã— 20

â‡’ Length of major axis = 40

Length of minor axis:

Length of minor axis = 2a (when a2 < b2)

= 2 Ã— 10

â‡’ Length of minor axis = 20

Eccentricity:

Eccentricity = c/b (when a2 < b2)

= 10âˆš3/20

= âˆš3/2

Length of the latus rectum:

Length of the latus rectum = 2a2/b (when a2 < b2)

= 2Ã—100/20

= 10

### Question 7. 36x2 + 4y2 = 144

Solution:

36x2 + 4y2 = 144

Dividing LHS and RHS by144,

= 1 (Obtained Equation)

Since denominator of y2/36 is larger than the denominator of x2/4,

the major axis is along the y-axis.

Comparing the given equation with  = 1, we get

a2 = 4 and b2 = 36

â‡’ a = Â±2 and b = Â±6

The Foci:

Foci = (0, c) and (0, -c) when (a2 < b2)

c = âˆš(b2 – a2)      -(when a2 < b2)

c = âˆš(36 – 4)

c = âˆš32

c = 4âˆš2

â‡’ (0, 4âˆš2) and (0, -4âˆš2)

Vertices:

Vertices = (0, b) and (0, -b) when (a2 < b2)

â‡’ (0, 6) and (0, -6)

Length of major axis:

Length of major axis = 2b (when a2<b2)

= 2 Ã— 6

â‡’ Length of major axis = 12

Length of minor axis

Length of minor axis = 2a (when a2 < b2)

= 2 Ã— 2

â‡’ Length of minor axis = 4

Eccentricity:

Eccentricity = c/b (when a2 < b2)

= 4âˆš2/6

= 2âˆš2/3

Length of the latus rectum:

Length of the latus rectum = 2a2/b (when a2 < b2)

= 2 Ã— 4/6

= 4/3

### Question 8. 16x2 + y2 = 16

Solution:

16x2 + y2 = 16

Dividing LHS and RHS by16,

= 1 (Obtained Equation)

Since denominator of y2/16 is larger than the denominator of x2/1,

the major axis is along the y-axis.

Comparing the given equation with  = 1, we get

a2 = 1 and b2 = 16

â‡’ a = Â±1 and b = Â±4

The Foci:

Foci = (0, c) and (0, -c) when (a2 < b2)

c = âˆš(b2 – a2)       -(when a2 < b2)

c = âˆš(16 – 1)

c = âˆš15

â‡’ (0, âˆš15) and (0, -âˆš15)

Vertices:

Vertices = (0, b) and (0, -b) when (a2 < b2)

â‡’ (0, 4) and (0, -4)

Length of major axis:

Length of major axis = 2b (when a2 < b2)

= 2 Ã— 4

â‡’ Length of major axis = 8

Length of minor axis:

Length of minor axis = 2a (when a2 < b2)

=2 Ã— 1

â‡’ Length of minor axis = 2

Eccentricity:

Eccentricity = c/b (when a2 < b2)

= âˆš15/4

Length of the latus rectum:

Length of the latus rectum = 2a2/b (when a2 < b2)

= 2 Ã— 1/4

= 1/2

### Question 9. 4x2 + 9y2 = 36

Solution:

4x2 + 9y2 = 36

Dividing LHS and RHS by 36,

= 1 (Obtained Equation)

Since denominator of x2/9 is larger than the denominator of y2/4,

the major axis is along the x-axis.

Comparing the given equation with  = 1, we get

a2 = 9 and b2 = 4

â‡’ a = Â±3 and b = Â±2

The Foci:

Foci = (c, 0) and (-c, 0) when (a2 > b2)

c = âˆš(a2 – b2)      -(when a2 > b2)

c = âˆš(9 – 4)

c = âˆš5

â‡’ (âˆš5, 0) and (-âˆš5, 0).

Vertices

Vertices = (a, 0) and (-a, 0) when (a2 > b2)

â‡’ (3, 0) and (-3, 0).

Length of major axis

Length of major axis = 2a (when a2 > b2)

= 2 Ã— 3

â‡’ Length of major axis = 6

Length of minor axis

Length of minor axis = 2b (when a2 > b2)

= 2 Ã— 2

â‡’ Length of minor axis = 4

Eccentricity

Eccentricity = c/a (when a2 > b2)

= âˆš5/3

Length of the latus rectum

Length of the latus rectum = 2b2/a (when a2 > b2)

= 2 Ã— 4/3

= 8/3

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

Solution:

Since the vertices are on x-axis, the equation will be of the form

= 1, where a is the semi-major axis. (where a2 > b2)

Given that a = Â±5, c = Â±4

As, from the relation

c2 = a2 – b2 (when a2 > b2)

b2 = a2 – c2

b2 = 25 – 16

b2 = 9

So, a2 = 25 and b2 = 9

Hence, the required equation of ellipse,

= 1

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

Solution:

Since the vertices are on y-axis, the equation will be of the form

= 1, where b is the semi-major axis. (where a2 < b2)

Given that b = Â±13, c = Â±5

As, from the relation

c2 = b2 – a2 (when a2 < b2)

a2 = b2 – c2

a2 = 169 – 25

a2 = 144

So, a2 = 144 and b2 = 169

Hence, the required equation of ellipse,

= 1

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

Solution:

Since the vertices are on x-axis, the equation will be of the form

= 1, where a is the semi-major axis. (where a2 > b2)

Given that a = Â±6, c = Â±4

As, from the relation

c2 = a2 – b2 (when a2 > b2)

b2 = a2 – c2

b2 = 36 – 16

b2 = 20

So, a2 = 36 and b2 = 20

Hence, the required equation of ellipse,

= 1

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

Solution:

Since the major axis are on x-axis, and minor axis on the y-axis, the equation will be of the form

= 1, where a is the semi-major axis. (where a2 > b2)

Given that a = Â±3, b = Â±2

So, a2 = 9 and b2 = 4

Hence, the required equation of ellipse,

= 1

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

Solution:

Since the major axis are on y-axis, and minor axis on the x-axis, the equation will be of the form

= 1, where b is the semi-major axis. (where a2 < b2)

Given that a = Â±1, b = Â±âˆš5

So, a2 = 1 and b2 = 5

Hence, the required equation of ellipse,

= 1

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

Solution:

Since the foci are on x-axis, the equation will be of the form

= 1, where a is the semi-major axis. (where a2 > b2)

Given that c = Â±5 and Length of major axis = 26

As, Length of major axis = 2a (when a2 > b2)

2a = 26

a = 13

As, from the relation

c2 = a2 – b2 (when a2 > b2)

b2 = a2 – c2

b2 = 169 – 25

b2 = 144

So, a2 = 169 and b2 = 144

Hence, the required equation of ellipse,

= 1

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

Solution:

Since the foci are on y-axis, the equation will be of the form

= 1, where b is the semi-major axis. (where a2 < b2)

Given that c = Â±6 and Length of minor axis = 16

As, Length of minor axis = 2a (when a2 < b2)

2a = 16

a = 8

As, from the relation

c2 = b2 – a2 (when a2 < b2)

b2 = c2 + a2

b2 = 36 + 64

b2 = 100

So, a2 = 64 and b2 = 100

Hence, the required equation of ellipse,

= 1

### Question 17. Foci (Â± 3, 0), a = 4.

Solution:

Since the foci are on x-axis, the equation will be of the form

= 1, where a is the semi-major axis. (where a2 > b2)

Given that a = 4 and c = Â±3

As, from the relation

c2 = a2 – b2 (when a2 > b2)

b2 = a2 – c2

b2 = 16 – 9

b2 = 7

So, a2 = 16 and b2 = 7

Hence, the required equation of ellipse,

= 1

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

Solution:

Since the foci are on x-axis, the equation will be of the form

= 1, where a is the semi-major axis. (where a2 > b2)

Given that b = 3 and c = 4

As, from the relation

c2 = a2 – b2 (when a2 > b2)

a2 = b2 + c2

a2 = 9 + 16

a2 = 25

So, a2 = 25 and b2 = 9

Hence, the required equation of ellipse,

= 1

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

Solution:

The standard equation of ellipse having centre (0, 0) will be of the form

= 1

Since the points (3, 2) and (1, 6) lie on the ellipse, we can have

= 1

= 1         -(1)

= 1

= 1        -(2)

Eq(2) subtracted from (multiplying eq(1) by 9) and, we get

9Ã—() – () = 9 – 1

= 8

80/a2 = 8

a2 = 80/8

a2 = 10

Now, substituting a2 = 10 in eq(1)

= 1

9/10 + 4/b2 = 1

4/b2 = 1 – 9/10

4/b2 = 1/10

b2 = 10 Ã— 4 = 40

So, a2 = 10 and b2 = 40

Hence, the required equation of ellipse,

= 1

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

Solution:

The standard equation of ellipse having centre (0, 0) will be of the form

= 1

Since the points (4,3) and (6,2) lie on the ellipse, we can have

= 1

= 1         -(1)

and,  = 1

= 1         -(2)

(multiplying eq(2) by 9) subtracted from (multiplying eq(1) by 4) and, we get

9Ã—( ) – 4Ã—() = 9 – 4

= 5

260/a2 = 5

a2 = 260/5

a2 = 52

Now, substituting a2 = 52 in eq(1)

= 1

9/b2 = 1 – 16/52

9/b2 = 36/52

9/b2 = 36/52

b2 = 9 Ã— 36/52 = 13

So, a2 = 52 and b2 = 13

Hence, the required equation of ellipse,

= 1

Previous
Next