Q. The distance between the foci of the hyperbola $x^{2} - 3y^{2} - 4x - 6y - 11 = 0$ is
A) 4
B) 6
C) 8
D) 10
Explanation
(c)
Given equation can be rewritten as
$\frac{(x - 2)^{2}}{12} - \frac{(y - 1)^{2}}{4} = 1$
Now, $e = \sqrt{1 + \frac{4}{14}} = \frac{2}{\sqrt{3}}$
$\therefore$ Distance between foci $= 2ae = 2 \times \sqrt{12} \times \frac{2}{\sqrt{3}} = 8$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.