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

Chapter 11 Conic Sections Exercise 11.4 Class 11 Maths NCERT Solutions is very helpful guide in boosting your marks in examinations. NCERT Solutions for Class 11 Maths will help in covering the important concepts related to syllabus through which you can solve difficult questions easily.

1. Find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbola x

Comparing the equation x

we have, a = 4, b = 3

and c = √(a

Therefore, the coordinates of the foci are (± 5, 0) and that of vertices are (± 4, 0).

Also, the eccentricity e = c/a = 5/4 and the length of latus rectum = 2b

2. Find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbola

y

Comparing the equation y

we have, a = 3, b = 3√3

and c = √(a

Therefore, the coordinates of the foci are (0, ± 6) and that of vertices are (0, ± 3). Also, the eccentricity e = c/a = 5/4 and the length of latus rectum = 2b

3. Find the coordinates of the foci and the vertices, the eccentricity and length of the latus rectum of the hyperbola.

9y^{2}/16 – y^{2}/9 = 1.**Answer**Comparing the equation x

^{2}/16 – y^{2}/9 = 1 with the standard equation x^{2}/a^{2}– y^{2}/b^{2}= 1,we have, a = 4, b = 3

and c = √(a

^{2}+ b^{2}) = √(16 + 9) = √25 = 5Therefore, the coordinates of the foci are (± 5, 0) and that of vertices are (± 4, 0).

Also, the eccentricity e = c/a = 5/4 and the length of latus rectum = 2b

^{2}/a = 2(3)^{2}/4 = 9/2 units.2. Find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbola

y

^{2}/9 – x^{2}/27 = 1**Answer**Comparing the equation y

^{2}/9 – x^{2}/27 = 1 with the standard equation.we have, a = 3, b = 3√3

and c = √(a

^{2}+ b^{2}) = √(9 + 27) = √36 = √6Therefore, the coordinates of the foci are (0, ± 6) and that of vertices are (0, ± 3). Also, the eccentricity e = c/a = 5/4 and the length of latus rectum = 2b

^{2}/a = 2(3√3)^{2}/3 = 54/3 = 18 units.3. Find the coordinates of the foci and the vertices, the eccentricity and length of the latus rectum of the hyperbola.

^{ 2}– 4x

^{ 2}= 36

**Answer**

9y

^{ 2}– 4x

^{ 2}= 36 is the equation of hyperbola

i.e., y

^{2}/4 – x

^{2}/9 = 1

∴ a

^{ 2 }= 4, b

^{ 2 }= 9, ∴ c

^{ 2 }= a

^{ 2 }+ b

^{ 2 }= 4 + 9 = 13,

a = 2, b = 3, c = √3

Axis is y-axis

Foci (0, ± √13), vertices = (0, ± 2)

Eccentricity = e = c/a = √13/2 ,

Latus rectum = 2b

^{2}/a = (2 × 9)/2 = 9

4. Find the coordinates of the foci and the vertices, the eccentricity and length of the latus rectum of the hyperbola

16x

^{ 2 }– 9y

^{2}= 576

**Answer**

Equation of hyperbola is

16x

^{ 2}– 9y

^{ 2}= 576

or x

^{2}/36 – y

^{2}/64 = 1

∴ a

^{ 2 }= 36, b

^{ 2 }= 64, c

^{ 2 }= 36 + 64 = 100

∴ a = 6, b = 8, c = 10

Axis is x-axis

Foci are (± 10, 0)

Vertices are (± 6, 0)

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

Latus rectum = 2b

^{2}/a = (2 × 64)/6 = 64/3

5. Find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbolas

5y

^{ 2 }– 9x

^{ 2}= 36

**Answer**

Equation of hyperbolas is 5y

^{ 2 }– 9x

^{ 2}= 36

=> y

^{2}/36/5 - x

^{2}/4 = 1 => -x

^{2}/4 + y

^{2}/36/5 = 1

∴ b

^{2}= 36/5 , a

^{ 2}= 4, c

^{ 2}= b

^{ 2}+ a

^{ 2}

= 4 + 36/5 = 56/5

∴ b = 6/√5 , a = 2, c = 2√14/√5

Axis is along y-axis ∴Foci are (0, ± 2√14/√5) ,

vertices are (0, ± 6/√5)

Eccentricity, e = c/b = 2√14/√5 × √5/6 = √14/3;

Latus rectum = 2a

^{2}/b = 2 × 4/6/√5 = 4√5/3

6. Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola

49y

^{ 2}– 16x

^{ 2}= 784

**Answer**

Dividing the equation by 784 on both sides, we have

49y

^{2}/a – 16x

^{2}/784 = 1 => y

^{2}/16 – x

^{2}/49 = 1

Comparing the equation with the standard equation y

^{2}/a

^{2}– x

^{2}/b

^{2}= 1 we find that a = 4, b = 7 and

c = √(a

^{2 }+ b

^{2 }) = √( 16 + 49) = √65

Therefore the coordinates of the foci are (0, ± √65) and that of vertices are (0, ± 4). Also the eccentricity e = c/a = √65/4 and the length of latus rectum is 2b

^{2}/a = 2(7)

^{2}/4 = 49/2 units.

7. Find the equation of the hyperbola satisfying the given conditions:

Vertices (± 2, 0), foci (± 3, 0)

**Answer**

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

x

^{2}/a

^{2}– y

^{2}/b

^{2}= 1

Given: vertices are (± 2, 0), a = 2

Also, since foci are (± 3, 0), c = 3 and

b

^{2}= c

^{ 2 }– a

^{2}= 9 – 4 = 5

Therefore, the equation of the hyperbola is

x

^{2}/4 – y

^{2}/5 = 1

8. Find the equation of the hyperbola satisfying the given conditions: Vertices (0, ± 5), foci (0, ± 8)

**Answer**

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

y

^{2}/a

^{2}– x

^{2}/b

^{2}= 1

Given: vertices are (0, ± 5), a = 5

Also, since foci are (0, ± 8), c = 8 and

b

^{ 2 }= c

^{ 2 }– a

^{ 2 }= 64 – 25 = 39

Therefore, the equation of the hyperbola is

x

^{2}/25 – y

^{2}/39 = 1

9. Find the equation of the hyperbola satisfying the given conditions:

Vertices (0, ± 3), foci (0, ± 5)

**Answer**

Vertices are (0, ± 3) ∴ a = 3

Foci are (0, ±5)

∴ c = 5, b

^{ 2 }= c

^{ 2 }– a

^{ 2 }= 25 – 9 = 16

∴ b = 4

∴ Equation of hyperbola is (axis being y-axis)

y

^{2}/9 – x

^{2}/16 = 1

10. Find the equation of the hyperbola satisfying the given conditions:

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

**Answer**

Foci are (± 5, 0) ∴ c = 5

Transverse axis = 8 ∴ a = 4

b

^{ 2 }= c

^{ 2}– a

^{ 2 }= 25 – 16 = 9 ∴ b = 3

Axis of hyperbola is x-axis since (± 5, 0) lies on it.

x

^{2}/16 – y

^{2}/9 = 1

11. Find the equation of the hyperbola satisfying the given conditions:

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

**Answer**

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

y

^{2}/a

^{2}– x

^{2}/b

^{2}= 1

Given: foci are (0, ± 13), c = 13

Length of conjugate axis = 2b = 24 => b = 12

or b

^{ 2}= c

^{ 2}– a

^{ 2}or 144 = 169 – a

^{ 2}

or a

^{ 2}= 25 or a = 5

The equation of the hyperbola is

y

^{2}/25 – x

^{2}/144 = 1

or 144 y

^{ 2}– 25x

^{ 2}= 3600

12. Find the equation of the hyperbola satisfying the given conditions:

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

**Answer**

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

x

^{2}/a

^{2}– y

^{2}/b

^{2}= 1

Given: foci are (± 3√5, 0), c = 3√5

and length of latus rectum = 2b

^{2}/a = 8

As b

^{ 2}= 4a

We have c

^{ 2}= a

^{ 2}+ b

^{ 2}

or 45 = a

^{ 2}+ 4 a

or a

^{ 2}+ 4 a – 45 = 0

or a

^{ 2}+ 9a – 5a – 45 = 0

or a(a + 9) – 5 (a – 9) = 0

or (a + 9)(a – 5) = 0

or a = –9 or a = 5

Since a cannot be negative, we take a = 5 and so b

^{ 2}= 20

Therefore, the equation of the required hyperbola is x

^{2}/25 – y

^{2}/20 = 1 => 4x

^{2}– 5y

^{2}= 100

13. Find the equation of hyperbola satisfying the given condition:

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

**Answer**

Foci are (± 4, 0) ∴ c = 4

or c

^{2}= a

^{ 2}+ b

^{ 2}∴ 16 = a

^{ 2 }+ b

^{ 2 }… (i)

Latus rectum = 2b

^{2}/a = 12

∴ b

^{ 2 }= 6a … (ii)

Eliminating b

^{ 2 }from (i) and (ii)

∴ 16 = a

^{ 2 }+ 6a or a

^{ 2 }+ 6a – 16 = 0

or (a + 8) (a – 2) = 0

a

^{2}– 8, ∴ a = 2 ∴ b

^{ 2 }= 6a = 6 × 2 = 12

a

^{ 2 }= 4, b

^{ 2 }= 12, Axis is x-axis

x

^{2}/4 – y

^{2}/12 = 1

14. Find the equation of hyperbola satisfying the given condition:

Vertices (± 4, 0), e = 4/3

**Answer**

Veritices are (± 7, 0) ∴ a = 7

e = c/a = 4/3 ∴ c = 4/3 a = 4/3 × 7 = 28/3

b

^{2}= c

^{2}– a

^{2}= (28/3)

^{2}– 49 = 343/9

Axis is along x-axis

x

^{2}/49 - y

^{2}/343/9 = 1 or 7x

^{2}– 9y

^{2}= 343

15. Find the equation of the hyperbola satisfying the given condition:

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

**Answer**

Let the equation of hyperbola be

y

^{2}/b

^{2}- x

^{2}/a

^{2}= 1 …(i)

∴ be = √10

Also, a

^{ 2}= b

^{ 2}(e

^{ 2}– 1) = b

^{ 2}e

^{ 2}– b

^{ 2}= 10 – b

^{ 2}…(ii)

Thus, the equation of the hyperbola is

y

^{2}/b

^{2}- x

^{2}/(10 – b

^{2}) = 1

As, it passes through the point (2, 3)

∴ 9/b

^{2}- 4/(10 – b

^{2}) = 1 => b

^{ 4}– 23b

^{ 2}+ 90 = 0

=> (b

^{ 2}– 5)(b

^{ 2}– 18) = 0 => b

^{ 2}= 18, 5

When b

^{ 2}= 18, then from (ii) a

^{ 2}= –8 which is not possible and when b

^{ 2}= 5, then from (ii) a

^{ 2}= 5

Hence, the required equation of the hyperbola is y

^{2}/5 - x

^{2}/5 = 1 => y

^{2}- x

^{2}= 5