Q. If$\ e_{1}\$and$\ e_{2}$ are the eccentricities of a hyperbola $3x^{2} - 3y^{2} = 25$ and its conjugate, then
A) $e_{1}^{2} + \ e_{2}^{2} = 2$
B) $e_{1}^{2} + \ e_{2}^{2}\ = 4$
C) $e_{1} + e_{2}\ = 4$
D) $e_{1} + e_{2}\ = \sqrt{2}$
Explanation
(b)
Given, $x^{2} - y^{2} = \frac{25}{3}$
$\therefore\ e_{1} = \sqrt{1 + \frac{b^{2}}{a^{2}}} = \ \sqrt{1 + 1} = \sqrt{2}$
The equation of conjugate hyperbola is
$- x^{2}\ + y^{2}\ = \frac{25}{3}$
$\therefore\ e_{2} = \ \sqrt{1 + \frac{b^{2}}{a^{2}}} = \sqrt{1 + 1} = \sqrt{2}$
$\therefore\ e_{1}^{2} + e_{2}^{2} = \left( \sqrt{2} \right)^{2} + \ \left( \sqrt{2} \right)^{2} = 4$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.