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.