Question
Question: The equation of the hyperbola whose foci are (6, 4) and (–4, 4) and eccentricity 2 is-...
The equation of the hyperbola whose foci are (6, 4) and (–4, 4) and eccentricity 2 is-
A
12x2 –4y2 – 24x + 32y – 127 = 0
B
12x2 + 4y2 + 24x – 32y – 127 = 0
C
12x2 – 4y2 – 24x – 32y + 127 = 0
D
12x2 – 4y2 + 24x + 32y + 127 = 0
Answer
12x2 –4y2 – 24x + 32y – 127 = 0
Explanation
Solution
Foci S1(6, 4) S2(–4, 4)
centre C ŗ (1, 4)
S1S2 = 10, 2ae = 10 Ž ae = 5 Ž a.2 = 5
Ž a = 5/2
b2 = a2(e2 – 1) = 25/4 (4 – 1) = 75/4
Equation (25/4)(x−1)2 – (75/4)(y−4)2 = 1