Q. The eccentricity of the hyperbola $\frac{x^{2}}{16} - \frac{y^{2}}{25} = 1$ is
A) $\frac{3}{4}$
B) $\frac{3}{5}$
C) $\frac{\sqrt{41}}{4}$
D) $\frac{\sqrt{41}}{5}$
Explanation
(c)
Given equation is $\frac{x^{2}}{16} - \frac{y^{2}}{25} = 1$
$\therefore\ e = \ \sqrt{1 + \frac{b^{2}}{a^{2}}} = \sqrt{1 + \frac{25}{16}} = \frac{\sqrt{41}}{4}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.