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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 (x1)2(25/4)\frac{(x - 1)^{2}}{(25/4)}(y4)2(75/4)\frac{(y - 4)^{2}}{(75/4)} = 1