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• NCERT Solutions for Class 11 Maths

# Class 11 NCERT Solutions- Chapter 11 Conic Section – Exercise 11.4

• Difficulty Level : Hard
• Last Updated : 09 Mar, 2021

### Question 1. = 1

Solution:

Comparing the given equation with  = 1

we conclude that transverse axis is along x-axis.

a2 = 16 and b2 = 9

a = ±4 and b = ±3

Foci:

Foci = (c, 0) and (-c, 0)

c = √(a2+b2)

c = √(16+9)

c = √25

c = 5

So the foci is (5, 0) and (-5, 0)

Vertices:

Vertices = (a, 0) and (-a, 0)

So the vertices is (4, 0) and (-4, 0)

Eccentricity:

Eccentricity = c/a = 5/4

Length of the latus rectum:

Length of the latus rectum = 2b2/a

= 2×9/4

= 9/2

### Question 2. = 1

Solution:

Comparing the given equation with  = 1

we conclude that transverse axis is along y-axis.

a2 = 9 and b2 = 27

a = ±3 and b = ±3√3

Foci:

Foci = (0, c) and (0, -c)

c = √(a2 + b2)

c = √(9 + 27)

c = √36

c = 6

So the foci is (0,6) and (0,-6)

Vertices:

Vertices = (0,a) and (0,-a)

So the vertices is (0,3) and (0,-3)

Eccentricity:

Eccentricity = c/a

= 6/3

= 2

Length of the latus rectum:

Length of the latus rectum = 2b2/a

= 2×27/3

= 18

### Question 3. 9y2 – 4x2 = 36

Solution:

9y2 – 4x2 = 36

On dividing LHS and RHS by 36,

9y2/36 – 4x2/36 = 36/36

= 1

Comparing the given equation with = 1

we conclude that transverse axis is along y-axis.

a2 = 4 and b2 = 9

a = ±2 and b = ±3

Foci:

Foci = (0, c) and (0, -c)

c = √(a2 + b2)

c = √(4 + 9)

c = √13

So the foci is (0, √13) and (0, -√13)

Vertices:

Vertices = (0, a) and (0, -a)

So the vertices is (0, 2) and (0, -2)

Eccentricity:

Eccentricity = c/a

= √13/2

Length of the latus rectum:

Length of the latus rectum = 2b2/a

= 2×9/2

= 9

### Question 4. 16x2 – 9y2 = 576

Solution:

16x2 – 9y2 = 576

On dividing LHS and RHS by 576,

16x2/576 – 9y2/576 = 576/576

= 1

Comparing the given equation with = 1

we conclude that transverse axis is along x-axis.

a2 = 36 and b2 = 64

a = ±6 and b = ±8

Foci:

Foci = (c,0) and (-c,0)

c = √(a2 + b2)

c = √(36 + 64)

c = √100

c = 10

So the foci is (10, 0) and (-10, 0)

Vertices:

Vertices = (a, 0) and (-a, 0)

So the vertices is (6, 0) and (-6, 0)

Eccentricity:

Eccentricity = c/a

= 10/6 = 5/3

Length of the latus rectum:

Length of the latus rectum = 2b2/a

= 2×64/6

= 64/3

### Question 5. 5y2 – 9x2 = 36

Solution:

5y2 – 9x2 = 36

On dividing LHS and RHS by 36,

5y2/36 – 9x2/36 = 36/36

Comparing the given equation with = 1,

we conclude that transverse axis is along y-axis.

a2 = 36/5 and b2 = 4

a = ±6/√5 and b = ±2

Foci:

Foci = (0, c) and (0, -c)

c = √(a2 + b2)

c = √(36/5 + 4)

c = √56/5

c = 2√14/√5

So the foci is (0, 2√14/√5) and (0, -2√14/√5)

Vertices:

Vertices = (0, a) and (0, -a)

So the vertices is (0,6/√5) and (0,-6/√5)

Eccentricity:

Eccentricity = c/a

= (2√14/√5)/6/√5

= √14/3

Length of the latus rectum:

Length of the latus rectum = 2b2/a

= 2×4/(6/√5)

= 4√5/3

### Question 6. 49y2 – 16x2 = 784

Solution:

49y2 – 16x2 = 784

On dividing LHS and RHS by 784, we get

49y2/784 – 16x2/784 = 784/784

Comparing the given equation with = 1

we conclude that transverse axis is along y-axis.

a2 = 16 and b2 = 49

a = ±4 and b = ±7

Foci:

Foci = (0, c) and (0, -c)

c = √(a2 + b2)

c = √(16 + 49)

c = √65

So the foci is (0, √65) and (0, -√65)

Vertices:

Vertices = (0, a) and (0, -a)

So the vertices is (0, 4) and (0, -4)

Eccentricity:

Eccentricity = c/a = √65/4

Length of the latus rectum:

Length of the latus rectum = 2b2/a

= 2×49/4

= 49/2

### Question 7. Vertices (± 2, 0), foci (± 3, 0).

Solution:

Since the foci is on x-axis, the equation of the hyperbola is of the form

= 1

As, Vertices (± 2, 0) and foci (±3, 0)

So, a = ±2 and c = ±3

As, c = √(a2 + b2)

b2 = c2 – a2

b2 = 9 – 4

b2 = 5

So, a2 = 4 and b2 = 5

Hence, the equation is

= 1

### Question 8. Vertices (0, ± 5), foci (0, ± 8).

Solution:

Since the foci is on y-axis, the equation of the hyperbola is of the form

= 1

As, Vertices (0, ±5) and foci (0, ±8)

So, a = ±5 and c = ±8

As, c = √(a2 + b2)

b2 = c2 – a2

b2 = 64 – 25

b2 = 39

So, a2 = 25 and b2 = 39

Hence, the equation is

= 1

### Question 9. Vertices (0, ± 3), foci (0, ± 5).

Solution:

Since the foci is on y-axis, the equation of the hyperbola is of the form

= 1

As, Vertices (0, ± 3) and foci (0, ± 5)

So, a = ±3 and c = ±5

As, c = √(a2 + b2)

b2 = c2 – a2

b2 = 25 – 9

b2 = 16

So, a2 = 9 and b2 = 16

Hence, the equation is

= 1

### Question 10. Foci (± 5, 0), the transverse axis is of length 8.

Solution:

Since the foci is on x-axis, the equation of the hyperbola is of the form

= 1

As, Foci (±5, 0) ⇒ c = ±5

Since, the length of the transverse axis is 8,

2a = 8

a = 8/2

a = 4

As, c = √(a2 + b2)

b2 = c2 – a2

b2 = 25 – 16

b2 = 9

So, a2 = 16 and b2 = 9

Hence, the equation is

= 1

### Question 11. Foci (0, ±13), the conjugate axis is of length 24.

Solution:

Since the foci is on y-axis, the equation of the hyperbola is of the form

= 1

As, Foci (0, ± 13) ⇒ c = ±13

Since, the length of the conjugate axis is 24,

2b = 24

b = 24/2

b = 12

As, c = √(a2 + b2)

a2 = c2 – b2

a2 = 169 – 144

a2 = 25

So, a2 = 25 and b2 = 144

Hence, the equation is

= 1

### Question 12. Foci (± 3√5, 0), the latus rectum is of length 8.

Solution:

Since the foci is on x-axis, the equation of the hyperbola is of the form

= 1

As, Foci (±3√5, 0) ⇒ c = ±3√5

Since, the length of latus rectum is 8,

2b2/a = 8

b2 = 8a/2

b2 = 4a          -(1)

As, c = √(a2 + b2)

b2 = 45 – a2

4a = 45 – a2

a2 + 4a – 45 = 0

a2 + 9a – 5a – 45 = 0

(a + 9)(a – 5) = 0

a ≠ -9 (a has to be positive due to eq(1))

Hence, a = 5

From eq(1), we get

b2 = 4(5)

b2 = 20

So, a2 = 25 and b2 = 20

Hence, the equation is

= 1

### Question 13. Foci (± 4, 0), the latus rectum is of length 12.

Solution:

Since the foci is on x-axis, the equation of the hyperbola is of the form

= 1

As, Foci (±4, 0) ⇒ c=±4

Since, the length of latus rectum is 12,

2b2/a = 12

b2 = 12a/2

b2 = 6a            -(1)

As, c = √(a2 + b2)

b2 = 16 – a2

6a = 16 – a2

a2 + 6a – 16 = 0

a2 + 8a – 2a – 16 = 0

(a + 8)(a – 2) = 0

a ≠ -8 (a has to be positive due to eq(1))

Hence, a = 2

From eq(1), we get

b2 = 6(2)

b2 = 12

So, a2 = 4 and b2 = 12

Hence, the equation is

= 1

### Question 14. Vertices (± 7, 0), e = 4/3.

Solution:

Since the vertex is on x-axis, the equation of the hyperbola is of the form

= 1

As, Vertices (±7, 0) ⇒ a = ±7

As e = 4/3

c/a = 4/3

c = 4a/3

c = 28/3

As, c = √(a2 + b2)

b2 = 784/9 – 49

b2 = 343/9

So, a2 = 49 and b2 = 343/9

Hence, the equation is

x2/49 – y2/(343/9) = 1

= 1

### Question 15. Foci (0, ±√10), passing through (2, 3).

Solution:

Since the foci is on y-axis, the equation of the hyperbola is of the form

= 1

As, Foci (0, ±√10) ⇒ c=±√10

As, c = √(a2 + b2)

b2 = c2 – a2

b2 = 10 – a2           -(1)

As (2, 3) passes through the curve, hence

32/a2 – 22/b2 = 1

9/a2 – 4/b2 = 1

9/a2 – 4/(10 – a2) = 1

9(10 – a2) – 4a2 = a2(10 – a2)

90  – 9a2 – 4a2 = 10a2 – a4

a4 – 23a2 + 90 = 0

a4 – 18a2 – 5a2 + 90 = 0

a2(a2 – 18) – 5(a2 – 18) = 0

(a2 – 18)(a2 – 5) = 0

a2 = 18 or 5

As, a < c in hyperbola

So a2 = 5

And, b2 = 10 – 5            -(From eq(1))

b2 = 5

So, a2 = 5 and b2 = 5

Hence, the equation is = 1

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