Chapter 11: Conic Sections Exercise 11.4

Given:

Foci (0, ±13) and the conjugate axis is of length 24.

Here, the foci are on the y-axis.

The equation of the hyperbola is of the form y²/a² – x²/b² = 1

Since the foci are (0, ±13), so, c = 13

Since the length of the conjugate axis is 24,

2b = 24

b = 24/2

= 12

It is known that a² + b² = c²

a² + 12² = 13²

a² = 169 – 144

= 25

∴ The equation of the hyperbola is y²/25 – x²/144 = 1

Given:

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

Here, the foci are on the x-axis.

The equation of the hyperbola is of the form x²/a² – y²/b² = 1

Since, the foci are (± 3√5, 0), so, c = ± 3√5

Length of the latus rectum is 8

2b²/a = 8

2b² = 8a

b² = 8a/2

= 4a

It is known that a² + b² = c²

a² + 4a = 45

a² + 4a – 45 = 0

a² + 9a – 5a – 45 = 0

(a + 9) (a -5) = 0

a = -9 or 5

Since a is non–negative, a = 5

So, b² = 4a

= 4 × 5

= 20

∴ The equation of the hyperbola is x²/25 – y²/20 = 1

Given:

Foci (± 4, 0) and the latus rectum is of length 12

Here, the foci are on the x-axis.

The equation of the hyperbola is of form x²/a² – y²/b² = 1

Since, the foci are (± 4, 0), so, c = 4

Length of the latus rectum is 12

2b²/a = 12

2b² = 12a

b² = 12a/2

= 6a

It is known that a² + b² = c²

a² + 6a = 16

a² + 6a – 16 = 0

a² + 8a – 2a – 16 = 0

(a + 8) (a – 2) = 0

a = -8 or 2

Since a is non–negative, a = 2

So, b² = 6a

= 6 × 2

= 12

∴ The equation of the hyperbola is x²/4 – y²/12 = 1

Given:

Vertices (±7, 0) and e = 4/3

Here, the vertices are on the x-axis

The equation of the hyperbola is of form x²/a² – y²/b² = 1

Since, the vertices are (± 7, 0), so, a = 7

It is given that e = 4/3

c/a = 4/3

3c = 4a

Substitute the value of a, and we get

3c = 4(7)

c = 28/3

It is known that a² + b² = c²

7² + b² = (28/3)²

b² = 784/9 – 49

= (784 – 441)/9

= 343/9

∴ The equation of the hyperbola is x²/49 – 9y²/343 = 1

Given:

Foci (0, ±√10) and passing through (2, 3)

Here, the foci are on the y-axis.

The equation of the hyperbola is of the form y²/a² – x²/b² = 1

Since, the foci are (±√10, 0), so, c = √10

It is known that a² + b² = c²

b² = 10 – a² ………….. (1)

It is given that the hyperbola passes through points (2, 3)

So, 9/a² – 4/b² = 1 … (2)

From equations (1) and (2), we get,

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

In hyperbola, c > a i.e., c² > a²

So, a² = 5

b² = 10 – a²

= 10 – 5

= 5

∴ The equation of the hyperbola is y²/5 – x²/5 = 1

Exercise 11.1 Solutions 15 Questions

Exercise 11.2 Solutions 12 Questions

Exercise 11.3 Solutions 20 Questions

Miscellaneous Exercise On Chapter 11 Solutions 8 Questions

Also explore – NCERT Class 11 Solutions