NCERT Solutions Class 11 Chapter 10 Conic Section Exercise 10.4 (original) (raw)

Last Updated : 22 Apr, 2024

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. \frac{x^2}{16} - \frac{y^2}{9} = 1

**Solution:

Comparing the given equation with\mathbf{\frac{x^2}{a^2} - \frac{y^2}{b^2}} = 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. \frac{y^2}{9} - \frac{x^2}{27} = 1

**Solution:

Comparing the given equation with\mathbf{\frac{y^2}{a^2} - \frac{x^2}{b^2}} = 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

\mathbf{\frac{y^2}{4} - \frac{x^2}{9}} = 1

Comparing the given equation with \mathbf{\frac{y^2}{a^2} - \frac{x^2}{b^2}} **= 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

\mathbf{\frac{x^2}{36} - \frac{y^2}{64}} = 1

Comparing the given equation with \mathbf{\frac{x^2}{a^2} - \frac{y^2}{b^2}} **= 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

\mathbf{\frac{y^2}{\frac{36}{5}} - \frac{x^2}{4}} = 1

Comparing the given equation with \mathbf{\frac{y^2}{a^2} - \frac{x^2}{b^2}} **= 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

\mathbf{\frac{y^2}{16} - \frac{x^2}{49}} = 1

Comparing the given equation with \mathbf{\frac{y^2}{a^2} - \frac{x^2}{b^2}} **= 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

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

\mathbf{\frac{x^2}{a^2} - \frac{y^2}{b^2}} **= 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

\mathbf{\frac{x^2}{4} - \frac{y^2}{5}} = 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

\mathbf{\frac{y^2}{a^2} - \frac{x^2}{b^2}} **= 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

\mathbf{\frac{y^2}{25} - \frac{x^2}{39}} **= 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

\mathbf{\frac{y^2}{a^2} - \frac{x^2}{b^2}} **= 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

\mathbf{\frac{y^2}{9} - \frac{x^2}{16}} **= 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

\mathbf{\frac{x^2}{a^2} - \frac{y^2}{b^2}} **= 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

\mathbf{\frac{x^2}{16} - \frac{y^2}{9}} **= 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

\mathbf{\frac{y^2}{a^2} - \frac{x^2}{b^2}} **= 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

\mathbf{\frac{y^2}{25} - \frac{x^2}{144}} **= 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

\mathbf{\frac{x^2}{a^2} - \frac{y^2}{b^2}} **= 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

\mathbf{\frac{x^2}{25} - \frac{y^2}{20}} **= 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

\mathbf{\frac{x^2}{a^2} - \frac{y^2}{b^2}} **= 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

\mathbf{\frac{x^2}{4} - \frac{y^2}{12}} **= 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

\mathbf{\frac{x^2}{a^2} - \frac{y^2}{b^2}} **= 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

\mathbf{\frac{x^2}{49} - \frac{9y^2}{343}} **= 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

\mathbf{\frac{y^2}{a^2} - \frac{x^2}{b^2}} **= 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 \mathbf{\frac{y^2}{5} - \frac{x^2}{5}} **= 1