Q. $P$ is any point on the ellipse $81x^{2} + 144y^{2} = 1944$, whose foci are $S$ and $S'$. Then, $SP + S^{'}P\$ equals
A) 3
B) $4\sqrt{6}$
C) 36
D) 324
Explanation
(b)
Given, $\frac{x^{2}}{24} + \frac{y^{2}}{13.5} = 1\$
$\therefore\ SP + S^{'}P = 2a = 4\sqrt{6}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.