Question -
Answer -
(i) the distance between the foci = 16 and eccentricity =√2
Given:
Distance between the foci = 16
Eccentricity = √2
Let us compare with the equation of the form
Distance between the foci is 2ae and b2 = a2(e2 –1)
So,
2ae = 16
ae = 16/2
a√2 = 8
a = 8/√2
a2 = 64/2
= 32
We know that, b2 = a2(e2 –1)
So, b2 = 32 [(√2)2 – 1]
= 32 (2 – 1)
= 32
The Equation of hyperbola is given as
x2 – y2 = 32
∴ The Equation of hyperbola is x2 – y2 =32
(ii) conjugate axis is 5 and the distance between foci = 13
Given:
Conjugate axis = 5
Distance between foci = 13
Let us compare with the equation of the form
Distance between the foci is 2ae and b2 = a2(e2 –1)
Length of conjugate axis is 2b
So,
2b = 5
b = 5/2
b2 = 25/4
We know that, 2ae = 13
ae = 13/2
a2e2 = 169/4
b2 = a2(e2 – 1)
b2 = a2e2 – a2
25/4 = 169/4 – a2
a2 = 169/4 – 25/4
= 144/4
= 36
The Equation of hyperbola is given as
∴ The Equation of hyperbola is 25x2 – 144y2 =900
(iii) conjugate axis is 7 and passes through the point (3,-2)
Given:
Conjugate axis = 7
Passes through the point (3, -2)
Conjugate axis is 2b
So,
2b = 7
b = 7/2
b2 = 49/4
The Equation of hyperbola is given as
Since it passes through points (3, -2)
∴ The Equation of hyperbola is 65x2 – 36y2 =441