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 x2/16 – y2/9 = 1.

Answer

Comparing the equation x2/16 – y2/9 = 1 with the standard equation x2/a2 – y2/b2 = 1,
we have, a = 4, b = 3
and c = √(a2 + b2) = √(16 + 9) = √25 = 5
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 = 2b2/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
y2/9 – x2/27 = 1

Answer

Comparing the equation y2/9 – x2/27 = 1 with the standard equation.
we have, a = 3, b = 3√3
and c = √(a2 + b2) = √(9 + 27) = √36 = √6
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 = 2b2/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.
9y 2 – 4x 2 = 36

Answer

9y 2 – 4x 2 = 36 is the equation of hyperbola
i.e., y2/4 – x2/9 = 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 = 2b2/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 – 9y2 = 576

Answer

Equation of hyperbola is
16x 2 – 9y 2 = 576
or x2/36 – y2/64 = 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 = 2b2/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
=> y2/36/5 - x2/4 = 1 => -x2/4 + y2/36/5 = 1
∴ b2 = 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 = 2a2/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
49y2/a – 16x2/784 = 1 => y2/16 – x2/49 = 1
Comparing the equation with the standard equation y2/a2 – x2/b2 = 1 we find that a = 4, b = 7 and
c = √(a+ b) = √( 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 2b2/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
x2/a2 – y2/b2 = 1
Given: vertices are (± 2, 0), a = 2
Also, since foci are (± 3, 0), c = 3 and
2 = c 2 – a 2 = 9 – 4 = 5
Therefore, the equation of the hyperbola is
x2/4 – y2/5 = 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
y2/a2 – x2/b2 = 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
x2/25 – y2/39 = 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)
y2/9 – x2/16 = 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.
x2/16 – y2/9 = 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
y2/a2 – x2/b2 = 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
y2/25 – x2/144 = 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
x2/a2 – y2/b2 = 1
Given: foci are (± 3√5, 0), c = 3√5
and length of latus rectum = 2b2/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 x2/25 – y2/20 = 1 => 4x2 – 5y2 = 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 c2 = a 2 + b 2 ∴ 16 = a 2 + b 2 … (i)
Latus rectum = 2b2/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
a2 – 8, ∴ a = 2 ∴ b 2 = 6a = 6 × 2 = 12
a 2 = 4, b 2 = 12, Axis is x-axis
x2/4 – y2/12 = 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
b2 = c2 – a2 = (28/3)2 – 49 = 343/9
Axis is along x-axis
x2/49 - y2/343/9 = 1 or 7x2 – 9y2 = 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
y2/b2 - x2/a2 = 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
y2/b2 - x2/(10 – b2) = 1
As, it passes through the point (2, 3)
∴ 9/b2 - 4/(10 – b2) = 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 y2/5 - x2/5 = 1 => y2 - x2 = 5
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