Question

The equation of the hyperbola whose foci are (6, 4) and (– 4, 4) and eccentricity 2 is given by

• A. $12x^{2} - 4y^{2} - 24x + 32y - 127 = 0$
• B. $12x^{2} + 4y^{2} + 24x - 32y - 127 = 0$
• C. $12x^{2} - 4y^{2} - 24x - 32y + 127 = 0$
• D. $12x^{2} - 4y^{2} + 24x + 32y + 127 = 0$

Answer 1

Answer: (A)

$12x^{2} - 4y^{2} - 24x + 32y - 127 = 0$

Answer 2

Foci are (6, 4) and (– 4, 4) and $e = 2$.

$\therefore$ Centre is $\left( \frac{6 - 4}{2},\frac{4 + 4}{2} \right) = (1,4)$

So, $ae + 1 = 6$ ⇒ $ae = 5$ ⇒ $a = \frac{5}{2}$ and $b = \frac{5}{2}\sqrt{3}$

Hence, the required equation is $\frac{(x - 1)^{2}}{25/4} - \frac{(y - 4)^{2}}{(75/4)} = 1$ or

$12x^{2} - 4y^{2} - 24x + 32y - 127 = 0$