Q. Eccentricity of hyperbola $\frac{x^{2}}{k} + \frac{y^{2}}{k^{2}} = 1(k < 0)$ is
A) $\sqrt{1 + k}$
B) $\sqrt{1 - k}$
C) $\sqrt{1 + \frac{1}{k}}$
D) $\sqrt{1 - \frac{1}{k}}$
Explanation
(d)
Given equation can be rewritten as $\frac{y^{2}}{k^{2}} - \frac{x^{2}}{( - k)} = 1( - k > 0)$
$e^{2} = 1 + \frac{( - k)}{k^{2}} = 1 - \frac{k}{k^{2}}$
$\Rightarrow \ \ e = \sqrt{1 - \frac{1}{k}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.