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

### In each of the Exercises 1 to 6, find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbolas.

### Question 1. = 1

**Solution:**

Comparing the given equation with

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free classeswhich will definitely help them in making a wise career choice in the future.we conclude that transverse axis is

along x-axis.a

^{2 }= 16 and b^{2 }= 9a = ±4 and b = ±3

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

c = √(a

^{2}+b^{2})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 = 2b

^{2}/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.a

^{2 }= 9 and b^{2 }= 27a = ±3 and b = ±3√3

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

c = √(a

^{2 }+ b^{2})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 = 2b

^{2}/a= 2×27/3

= 18

### Question 3. 9y^{2} – 4x^{2} = 36

**Solution:**

9y

^{2}– 4x^{2}= 36On dividing LHS and RHS by 36,

9y

^{2}/36 – 4x^{2}/36 = 36/36= 1

Comparing the given equation with

= 1,we conclude that transverse axis is

along y-axis.a

^{2 }= 4 and b^{2 }= 9a = ±2 and b = ±3

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

c = √(a

^{2 }+ b^{2})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 = 2b

^{2}/a= 2×9/2

= 9

### Question 4. 16x^{2} – 9y^{2} = 576

**Solution:**

16x

^{2}– 9y^{2}= 576On dividing LHS and RHS by 576,

16x

^{2}/576 – 9y^{2}/576 = 576/576= 1

Comparing the given equation with

= 1,we conclude that transverse axis is

along x-axis.a

^{2 }= 36 and b^{2 }= 64a = ±6 and b = ±8

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

c = √(a

^{2 }+ b^{2})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 = 2b

^{2}/a= 2×64/6

= 64/3

### Question 5. 5y^{2} – 9x^{2} = 36

**Solution:**

5y

^{2}– 9x^{2}= 36On dividing LHS and RHS by 36,

5y

^{2}/36 – 9x^{2}/36 = 36/36Comparing the given equation with

= 1,we conclude that transverse axis is

along y-axis.a

^{2 }= 36/5 and b^{2 }= 4a = ±6/√5 and b = ±2

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

c = √(a

^{2 }+ b^{2})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 = 2b

^{2}/a= 2×4/(6/√5)

= 4√5/3

### Question 6. 49y^{2} – 16x^{2} = 784

**Solution:**

49y

^{2}– 16x^{2}= 784On dividing LHS and RHS by 784, we get

49y

^{2}/784 – 16x^{2}/784 = 784/784Comparing the given equation with

= 1,we conclude that transverse axis is

along y-axis.a

^{2 }= 16 and b^{2 }= 49a = ±4 and b = ±7

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

c = √(a

^{2 }+ b^{2})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 = 2b

^{2}/a= 2×49/4

= 49/2

### In each of the Exercises 7 to 15, find the equations of the hyperbola satisfying the given conditions.

### 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

= 1As, Vertices (± 2, 0) and foci (±3, 0)

So, a = ±2 and c = ±3

As, c = √(a

^{2 }+ b^{2})b

^{2}= c^{2 }– a^{2}b

^{2}= 9 – 4b

^{2}= 5So, a

^{2}= 4 and b^{2}= 5Hence, 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

= 1As, Vertices (0, ±5) and foci (0, ±8)

So, a = ±5 and c = ±8

As, c = √(a

^{2 }+ b^{2})b

^{2}= c^{2 }– a^{2}b

^{2}= 64 – 25b

^{2}= 39So, a

^{2}= 25 and b^{2}= 39Hence, 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

= 1As, Vertices (0, ± 3) and foci (0, ± 5)

So, a = ±3 and c = ±5

As, c = √(a

^{2 }+ b^{2})b

^{2}= c^{2 }– a^{2}b

^{2}= 25 – 9b

^{2}= 16So, a

^{2}= 9 and b^{2}= 16Hence, 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

= 1As, Foci (±5, 0) ⇒ c = ±5

Since, the length of the transverse axis is 8,

2a = 8

a = 8/2

a = 4

As, c = √(a

^{2 }+ b^{2})b

^{2}= c^{2 }– a^{2}b

^{2}= 25 – 16b

^{2}= 9So, a

^{2}= 16 and b^{2}= 9Hence, 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

= 1As, Foci (0, ± 13) ⇒ c = ±13

Since, the length of the conjugate axis is 24,

2b = 24

b = 24/2

b = 12

As, c = √(a

^{2 }+ b^{2})a

^{2}= c^{2 }– b^{2}a

^{2}= 169 – 144a

^{2}= 25So, a

^{2}= 25 and b^{2}= 144Hence, 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

= 1As, Foci (±3√5, 0) ⇒ c = ±3√5

Since, the length of latus rectum is 8,

2b

^{2}/a = 8b

^{2}= 8a/2b

^{2}= 4a -(1)As,

c = √(a^{2 }+ b^{2})b

^{2}= 45 – a^{2}4a = 45 – a

^{2}a

^{2 }+ 4a – 45 = 0a

^{2 }+ 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

b

^{2}= 4(5)b

^{2}= 20So, a

^{2}= 25 and b^{2}= 20Hence, 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

= 1As, Foci (±4, 0) ⇒ c=±4

Since, the length of latus rectum is 12,

2b

^{2}/a = 12b

^{2}= 12a/2b

^{2}= 6a -(1)As, c = √(a

^{2 }+ b^{2})b

^{2}= 16 – a^{2}6a = 16 – a

^{2}a

^{2 }+ 6a – 16 = 0a

^{2 }+ 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

b

^{2}= 6(2)b

^{2}= 12So, a

^{2}= 4 and b^{2}= 12Hence, 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

= 1As, Vertices (±7, 0) ⇒ a = ±7

As e = 4/3

c/a = 4/3

c = 4a/3

c = 28/3

As, c = √(a

^{2 }+ b^{2})b

^{2}= 784/9 – 49b

^{2}= 343/9So, a

^{2}= 49 and b^{2}= 343/9Hence, the equation is

x

^{2}/49 – y^{2}/(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

= 1As, Foci (0, ±√10) ⇒ c=±√10

As, c = √(a

^{2 }+ b^{2})b

^{2}= c^{2 }– a^{2}b

^{2}= 10 – a^{2 }-(1)As (2, 3) passes through the curve, hence

3

^{2}/a^{2}– 2^{2}/b^{2}= 19/a

^{2}– 4/b^{2}= 19/a

^{2}– 4/(10 – a^{2}) = 19(10 – a

^{2}) – 4a^{2}= a^{2}(10 – a^{2})90 – 9a

^{2}– 4a^{2}= 10a^{2}– a^{4}a

^{4}– 23a^{2}+ 90 = 0a

^{4}– 18a^{2}– 5a^{2}+ 90 = 0a

^{2}(a^{2}– 18) – 5(a^{2}– 18) = 0(a

^{2}– 18)(a^{2}– 5) = 0a

^{2}= 18 or 5As, a < c in hyperbola

So a

^{2}= 5And, b

^{2}= 10 – 5 -(From eq(1))b

^{2}= 5So, a

^{2}= 5 and b^{2 }= 5Hence, the equation is

= 1