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

### Conic Sections Exercise 11.4 Solutions

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

Solution

The given equation is
On comparing the equation with the standard equation of hyperbola i.e., x2 /a2 - y2 /b2 = 1, we obtain a = 4 and b = 3.
We know that a2 + b2 = c2.
∴ c2 = 42 + 32 = 25
⇒ c = 5
Therefore,
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 = (2×9)/4 = 9/2.

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

Solution

The given equation is
On comparing this equation with the standard equation of hyperbola i.e., y2/a2 - x2/b2 = 1, we obtain a = 3 and b = √27.
We know that a2 + b2 = c2.
∴ c2 = 32 + (√27)2 = 9 + 27 = 36
⇒ c = 6
Therefore,
The coordinates of the foci are (0, ± 6).
The coordinates of the vertices are (0, ± 3).
Eccentricity, e = c/a = 6/3 = 2
Length of latus rectum = 2b2/a = (2× 27)/3 = 18

3. Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola 9y2 – 4x2 = 36.

Solution

The given equation is 9y2 – 4x2 = 36.
It can be written as
9y2 – 4x2 = 36

On comparing equation (1) with the standard equation of hyperbola i.e., y2/a2 - x2/b2 = 1, we obtain a = 2 and b = 3.
We know that a2 + b2 = c2.
∴ c2 = 4 + 9 = 13
⇒ c = √13
Therefore,
The coordinates of the foci are (0, ± √13)
The coordinates of the vertices are (0, ± 2).
Eccentricity,  e = c/a = √13/2
Length of latus rectum  = 2b2/a = (2× 9)/2 = 9

4. Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola 16x2 – 9y2 = 576

Solution

The given equation is 16x2 – 9y2 = 576.
It can be written as
16x2 – 9y2 = 576

On comparing equation (1) with the standard equation of hyperbola i.e., x2/a2 - y2/b2 = 1 , we obtain a = 6 and b = 8.
We know that a2 + b2 = c2.
∴ c2 = 36 + 64 = 100
⇒ c = 10
Therefore,
The coordinates of the foci are (± 10, 0).
The coordinates of the vertices are ( ±6, 0).
Eccentricity, e = c/a = 10/6 = 5/3
Length of 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 hyperbola 5y2 – 9x2 = 36

Solution

The given equation is 5y2 - 9x2 = 36.

On comparing equation (1) with the standard equation of hyperbola i.e., y2/a2 - x2/b2 = 1, we obtain a = 6/√5 and b = 2.
We know that a2 + b2 = c2.

6. Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola 49y2 – 16x2 = 784

Solution

The given equation is 49y2 – 16x2 = 784.
It can be written as
49y2 – 16x2 = 784

On comparing equation (1) with the standard equation of hyperbola i.e., y2 /a2 - x2 /b2 = 1, we obtain a = 4 and b = 7
We know that a2 + b2 = c2.
∴ c2 = 16 + 49 = 65
⇒ c = √65
Therefore,
The coordinates of the foci are (0, ± √65).
The coordinates of the vertices are (0, ± 4).
Eccentricity, e = c/a = √65/4
Length of latus rectum  = 2b2/a = (2× 49)/4 = 49/2

7.  Find the equation of the hyperbola satisfying the give conditions: Vertices (±2, 0), foci (±3, 0).

Solution

Vertices (±2, 0), foci (±3, 0)
Here, the vertices are on the x-axis.
Therefore, the equation of the hyperbola is of the form x2/a2 - y2/b2  = 1.
Since the vertices are (±2, 0), a = 2.
Since the foci are (±3, 0), c = 3.
We know that a2 + b2 = c2.
∴ 22 + b2 = 32
b2 = 9 - 4 = 5
Thus, the equation of the hyperbola is x2/4 - y2 /5 = 1.

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

Solution

Vertices (0, ±5), foci (0, ±8)
Here, the vertices are on the y-axis.
Therefore, 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.
We know that a2 + b2 = c2
∴ 52 + b2 = 82
b2 = 64 - 25 = 39
Thus, the equation of the hyperbola is y2/25 - x2/39 = 1.

9. Find the equation of the hyperbola satisfying the give conditions: Vertices (0, ±3), foci (0, ±5) .

Solution

Vertices (0, ±3), foci (0, ±5)
Here, the vertices are on the y-axis.
Therefore, the equation of the hyperbola is of the form y2/a2 - x2 /b2  = 1.
Since the vertices are (0, ±3), a = 3.
Since the foci are (0, ±5), c = 5.
We know that a2 + b2 = c2.
∴32 + b2 = 52
⇒ b2 = 25 – 9 = 16
Thus, the equation of the hyperbola is y2/9 - x2/16 = 1.

10. Find the equation of the hyperbola satisfying the give conditions: Foci (±5, 0), the transverse axis is of length 8.

Solution

Foci (±5, 0), the transverse axis is of length 8.
Here, the foci are on the x-axis.
Therefore, the equation of the hyperbola is of the form x2/a2 - y2/b2  = 1.
Since the foci are (±5, 0), c = 5.
Since the length of the transverse axis is 8,
2a = 8
⇒ a = 4.
We know that a2 + b2 = c2.
∴ 42 + b2 = 52
⇒ b2 = 25 – 16 = 9
Thus, the equation of the hyperbola is x2/16 - y2/9 = 1.

11. Find the equation of the hyperbola satisfying the give conditions: Foci (0, ±13), the conjugate axis is of length 24.

Solution

Foci (0, ±13), the conjugate axis is of length 24.
Here, the foci are on the y-axis.
Therefore, the equation of the hyperbola is of the form y2/a2 - x2/b2 = 1.
Since the foci are (0, ±13), c = 13.
Since the length of the conjugate axis is 24,
2b = 24
⇒ b = 12.
We know that a2 + b2 = c2.
∴ a2 + 122 = 132
⇒ a2 = 169 – 144 = 25
Thus, the equation of the hyperbola is y2/25  - x2/144 = 1.

12. Find the equation of the hyperbola satisfying the give conditions: Foci (± 3√5, 0), the latus rectum is of length 8.

Solution

Foci (± 3√5, 0), the latus rectum is of length 8.
Here, the foci are on the x - axis.
Therefore, the equation of the hyperbola is of the form x2/a2 - y2/b2 = 1.
Since the foci are (±3√5, 0), c = ± 3√5 .
Length of latus rectum  = 8
⇒ 2b2 /a = 8
⇒ b2 = 4a
We know that a2 + b2 = c2 .
∴ a2 + 4a = 45
⇒ a2 + 4a – 45 = 0
⇒ a2 + 9a – 5a – 45 = 0
⇒ (a + 9) (a – 5) = 0
⇒⇒ a = –9, 5
Since a is non-negative, a = 5.
∴ b2 = 4a = 4 × 5 = 20
Thus, the equation of the hyperbola is x2/25 - y2/20 = 1.

13. Find the equation of the hyperbola satisfying the give conditions: Foci (±4, 0), the latus rectum is of length 12 .

Solution

Foci (±4, 0), the latus rectum is of length 12.
Here, the foci are on the x-axis.
Therefore, the equation of the hyperbola is of the form x2/a2 - y2/b2 = 1 .
Since the foci are (±4, 0), c = 4.
Length of latus rectum = 12
⇒ 2b2/a = 12
⇒ b2 = 6a
We know that a2 + b2 = c2.
∴ a2 + 6a = 16
⇒ a2 + 6a – 16 = 0
⇒ a2 + 8a – 2a – 16 = 0
⇒ (a + 8) (a – 2) = 0
⇒ a = –8, 2
Since a is non-negative, a = 2.
∴ b2 = 6a = 6×2 = 12
Thus, the equation of the hyperbola is x2/4 - y2/12 = 1.

14. Find the equation of the hyperbola satisfying the give conditions: Vertices (±7, 0), e = 4/3

Solution

Vertices (±7, 0), e = 4/3
Here, the vertices are on the x-axis.
Therefore, the equation of the hyperbola is of the form x2/a2 - y2/b2 = 1.
Since the vertices are (±7, 0), a = 7.
It is given that e = 4/3

Thus, the equation of the hyperbola is x2/49 - 9y2/343 = 1.

15. Find the equation of the hyperbola satisfying the give conditions: Foci (0, ±√10), passing through (2, 3) .

Solution

Foci (0, ±√10), passing through (2, 3)
Here, the foci are on the y-axis.
Therefore, the equation of the hyperbola is of the form y2/a2 - x2/b2 = 1.
Since the foci are (0, ± √10), c = √10 .
We know that a2 + b2 = c2.
∴ a2 + b2 = 10
⇒ b2 = 10 – a2 … (1)
Since the hyperbola passes through point (2, 3),
9/a2 - 4/b2 = 1  ...(2)
From equations (1) and (2), we obtain

⇒ 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
a2 = 5
b2 = 10 - a2 = 10 - 5 = 5
Thus, the equation of the hyperbola is y2/5 - x2/5 = 1.