Q. The equation of the ellipse whose foci are at $( \pm 2,\ 0)$ and eccentricity is $\frac{1}{2\ }$, is $\frac{x^{2}}{a^{2}\ }$ +$\frac{y^{2}}{b^{2}}$ =1. Then,
A) $a^{2} = 16,\ b^{2} = 12$
B) $a^{2} = 12,\ b^{2} = 16$
C) $a^{2} = 16,\ b^{2} = 4$
D) $a^{2} = 4,\ b^{2} = 16$
Explanation
(a)
Given , foci $( \pm ae,0) = ( \pm 2,0)$ and $e = \frac{1}{2}$
$\therefore\ ae = 2\ \Rightarrow \ a = 4$
Now, $b = a\sqrt{1 - \frac{1}{4}} = 2\sqrt{3}$
$\therefore\ a^{2} = 16\$and$\ b^{2} = 12$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.