NCERT Solutions for Class 11 Maths Chapter 11 - Conic Sections Exercise 11.4

*According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 10.

NCERT Solutions for Class 11 Maths Chapter 11 Exercise 11.4 have been carefully compiled and developed as per the latest update on the CBSE Syllabus. These solutions contain detailed step-by-step explanations of all the problems that are present in the NCERT textbook. Exercise 11.4 of NCERT Solutions for Class 11 Maths Chapter 11 – Conic Sections is based on the following topics:

1. Hyperbola
   1. Eccentricity
   2. Standard equation of Hyperbola
   3. Latus rectum

The solutions explain topics with engaging math activities that strengthen the concepts further. Each question of the exercise in NCERT Class 11 Maths Solutions has been carefully solved for the students to understand, keeping the board examinations in mind.

NCERT Solutions for Class 11 Maths Chapter 11 – Conic Sections Exercise 11.4

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Solutions for Class 11 Maths Chapter 11 – 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.

1. x²/16 – y²/9 = 1

Solution:

Given:

The equation is x²/16 – y²/9 = 1 or x²/4² – y²/3² = 1

On comparing this equation with the standard equation of hyperbola x²/a² – y²/b² = 1,

We get a = 4 and b = 3,

It is known that a² + b² = c²

So,

c² = 4² + 3²

= √25

c = 5

Then,

The coordinates of the foci are (±5, 0).

The coordinates of the vertices are (±4, 0).

Eccentricity, e = c/a = 5/4

Length of latus rectum = 2b²/a = (2 × 3²)/4 = (2×9)/4 = 18/4 = 9/2

2. y²/9 – x²/27 = 1

Solution:

Given:

The equation is y²/9 – x²/27 = 1 or y²/3² – x²/27² = 1

On comparing this equation with the standard equation of hyperbola y²/a² – x²/b² = 1,

We get a = 3 and b = √27,

It is known that a² + b² = c²

So,

c² = 3² + (√27)²

= 9 + 27

c² = 36

c = √36

= 6

Then,

The coordinates of the foci are (0, 6) and (0, -6).

The coordinates of the vertices are (0, 3) and (0, – 3).

Eccentricity, e = c/a = 6/3 = 2

Length of latus rectum = 2b²/a = (2 × 27)/3 = (54)/3 = 18

3. 9y² – 4x² = 36

Solution:

Given:

The equation is 9y² – 4x² = 36 or y²/4 – x²/9 = 1 or y²/2² – x²/3² = 1

On comparing this equation with the standard equation of hyperbola y²/a² – x²/b² = 1,

We get a = 2 and b = 3,

It is known that a² + b² = c²

So,

c² = 4 + 9

c² = 13

c = √13

Then,

The coordinates of the foci are (0, √13) and (0, –√13).

The coordinates of the vertices are (0, 2) and (0, – 2).

Eccentricity, e = c/a = √13/2

Length of latus rectum = 2b²/a = (2 × 3²)/2 = (2×9)/2 = 18/2 = 9

4. 16x² – 9y² = 576

Solution:

Given:

The equation is 16x² – 9y² = 576

Let us divide the whole equation by 576, we get

16x²/576 – 9y²/576 = 576/576

x²/36 – y²/64 = 1

On comparing this equation with the standard equation of hyperbola x²/a² – y²/b² = 1,

We get a = 6 and b = 8,

It is known that a² + b² = c²

So,

c² = 36 + 64

c² = √100

c = 10

Then,

The coordinates of the foci are (10, 0) and (-10, 0).

The coordinates of the vertices are (6, 0) and (-6, 0).

Eccentricity, e = c/a = 10/6 = 5/3

Length of latus rectum = 2b²/a = (2 × 8²)/6 = (2×64)/6 = 64/3

5. 5y² – 9x² = 36

Solution:

Given:

The equation is 5y² – 9x² = 36

Let us divide the whole equation by 36. We get

5y²/36 – 9x²/36 = 36/36

y²/(36/5) – x²/4 = 1

On comparing this equation with the standard equation of hyperbola y²/a² – x²/b² = 1,

We get a = 6/√5 and b = 2,

It is known that a² + b² = c²

So,

c² = 36/5 + 4

c² = 56/5

c = √(56/5)

= 2√14/√5

Then,

The coordinates of the foci are (0, 2√14/√5) and (0, – 2√14/√5).

The coordinates of the vertices are (0, 6/√5) and (0, -6/√5).

Eccentricity, e = c/a = (2√14/√5) / (6/√5) = √14/3

Length of latus rectum = 2b²/a = (2 × 2²)/6/√5 = (2×4)/6/√5 = 4√5/3

6. 49y² – 16x² = 784.

Solution:

Given:

The equation is 49y² – 16x² = 784.

Let us divide the whole equation by 784. We get

49y²/784 – 16x²/784 = 784/784

y²/16 – x²/49 = 1

On comparing this equation with the standard equation of hyperbola y²/a² – x²/b² = 1,

We get a = 4 and b = 7,

It is known that a² + b² = c²

So,

c² = 16 + 49

c² = 65

c = √65

Then,

The coordinates of the foci are (0, √65) and (0, –√65).

The coordinates of the vertices are (0, 4) and (0, -4).

Eccentricity, e = c/a = √65/4

Length of latus rectum = 2b²/a = (2 × 7²)/4 = (2×49)/4 = 49/2

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

7. Vertices (±2, 0), foci (±3, 0)

Solution:

Given:

Vertices (±2, 0) and foci (±3, 0)

Here, the vertices are on the x-axis.

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

Since, the vertices are (±2, 0), so, a = 2

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

It is known that a² + b² = c²

So, 2² + b² = 3²

b² = 9 – 4 = 5

∴ The equation of the hyperbola is x²/4 – y²/5 = 1

8. Vertices (0, ± 5), foci (0, ± 8)

Solution:

Given:

Vertices (0, ± 5) and foci (0, ± 8)

Here, the vertices are on the y-axis.

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

Since the vertices are (0, ±5), a = 5

Since the foci are (0, ±8), c = 8

It is known that a² + b² = c²

So, 5² + b² = 8²

b² = 64 – 25 = 39

∴ The equation of the hyperbola is y²/25 – x²/39 = 1

9. Vertices (0, ± 3), foci (0, ± 5)

Solution:

Given:

Vertices (0, ± 3) and foci (0, ± 5)

Here, the vertices are on the y-axis.

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

Since the vertices are (0, ±3), so, a = 3

Since the foci are (0, ±5), so, c = 5

It is known that a² + b² = c²

So, 3² + b² = 5²

b² = 25 – 9 = 16

∴ The equation of the hyperbola is y²/9 – x²/16 = 1

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

Solution:

Given:

Foci (±5, 0) and the transverse axis 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 (±5, 0), so, c = 5

Since the length of the transverse axis is 8,

2a = 8

a = 8/2

= 4

It is known that a² + b² = c²

4² + b² = 5²

b² = 25 – 16

= 9

∴ The equation of the hyperbola is x²/16 – y²/9 = 1

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

Solution:

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

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

Solution:

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

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

Solution:

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

14. Vertices (±7, 0), e = 4/3

Solution:

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

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

Solution:

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

Access Other Exercise Solutions of Class 11 Maths Chapter 11- Conic Sections

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