Q. If transverse and conjugate axes of hyperbola are equal then it’s eccentricity is
A) $\sqrt{3}$
B) $\sqrt{2}$
C) $\frac{1}{\sqrt{2}}$
D) 2
Explanation
(b)
Since, transverse and conjugate axes are equal
$ie,\ a = b\$
$\therefore\ \frac{x^{2}}{a^{2}} - \frac{y^{2}}{a^{2}} = 1$
Hence , $e = \sqrt{1 + \frac{a^{2}}{a^{2}}} = \sqrt{1 + 1} = \sqrt{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.