Q. If the foci of an ellipse are $( \pm \sqrt{5},\ 0)$ and its eccentricity is $\sqrt{5}/3$, then the equation of the ellipse is
A) $9x^{2} + 4y^{2} = 36$
B) $4x^{2} + 9y^{2} = 36$
C) $36x^{2} + 9y^{2} = 4$
D) $9x^{2} + 36y^{2} = 4$
Explanation
(b)
Given, eccentricity $e = \frac{\sqrt{5}}{3}$ and foci $= \left( \pm \ \sqrt{5,}o \right)$
$\Rightarrow \ ae = \ \sqrt{5}\ \Rightarrow \ a = 3$
$\therefore\ b^{2} = a^{2}\ (1 - e^{2}\_ = 9\left( 1 - \frac{5}{9} \right)$
$\Rightarrow \ b^{2} = 4$
The equation of ellipse is
$\frac{x^{2}}{9} + \frac{y^{2}}{4} = 1$
$\Rightarrow \ 4x^{2} + 9y^{2} = 36$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.