- 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 (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 i.e., , we obtain a = 4 and b = 3.
We know that a2 + b2 = c2
c2 = 42 + 32 = 16+ 9 = 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
- 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 (y2)/9 – (x2)/27 = 1 or (y2)/(32) – (x2)/(√27)2 = 1
On comparing this equation with the standard equation of hyperbola [(x2)/(a2)] – [(y2)/(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
- 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
or
(y2)/4 – (x2)/9 = 1
(y2)/(22) – (x2)/(32) = 1 ——–(1)
On comparing equation (1) with the standard equation of hyperbola [(x2)/(a2)] – [(y2)/(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
- 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
[(x2)/36] – [(y2)/64] = 1
[(x2)/62] – [(y2)/82] = 1 ————(1)
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
- 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.
(y2)/[36/5] – (x2)/4 = 1
(y2)/[6/√5]2 – (x2)/(22) = 1 —————(1)
On comparing equation (1) with the standard equation of hyperbola [(y2)/(a2)] – [(x2)/(b2)] = 1, we obtain a =6/√5 and b = 2.
We know that a2 + b2 = c2
c2 = 36/5 + 4 = 56/5
c = √(56/5) = 2√14/√5
Therefore, the coordinates of the foci are (0, ±2√14/√5)
The coordinates of the vertices are (0, ±6/√5)
Eccentricity, e = c/a = [2√14/√5]/[ 6/√5] = √14/3
Length of latus rectum = 2b2/a = 2×4/(6/√5) = 4x√5/3
- 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
(y2)/16 – (x2)/49 = 1
(y2)/(42) – (x2)/(72) = 1 ————(1)
On comparing equation (1) with the standard equation of hyperbola (y2)/(a2) – (x2)/(b2) = 1 i.e., , 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
- 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
- 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 .
- 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
- 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
b = 3
Thus, the equation of the hyperbola is .(x2)/16 – (y2)/9 = 1
- 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
- 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 or 5
Since a is non- negative, a = 5
b2 = 4a = 4 x 5 = 20
Thus, the equation of the hyperbola is x2/25 – y2/20 = 1
- 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 + 8a – 6a – 16 = 0
(a + 8)(a – 2) = 0
a = -8 or 2
Since a is non negative, a = 2
b2 = 6a = 6 x 2 = 12
Thus, the equation of the hyperbola is x2/4 – y2/12 = 1
- 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 , we know e = c/a
c/a = 4/3
c/7 = 4/3
c = 28/3
We know that a2 + b2 = c2.
72 + b2 = (28/3)2
b2 = 784/9 – 49
b2 = (784 – 441)/9 = 343/9
Thus the equation of the hyperbola x2/49 – (9y2)/343 = 1
- 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/a2 – 4/(10-a2) = 1
9(10 – a2) – 4a2 = a2(10 – a2)
90 – 9a2 – 4a2 = a210 – a4
a4– 23a2 + 90 = 0
a4– 18a2– 5a2 + 90 = 0
a2(a2 – 18)-5(a2 – 18) = 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