JEE Main 2020 (Online) 5th September Morning ShiftQuestion 1
If the co-ordinates of two points $A$ and $B$ are $(\sqrt7,0)$ and $(-\sqrt7,0)$ respectively and $P$ is any point on the conic, $9x^2+16y^2=144$, then $PA+PB$ is equal to :
Ellipse
Solution
$\frac{x^2}{16}+\frac{y^2}{9}=1$: $a=4$, $c^2=16-9=7$, so foci are $(\pm\sqrt7,0)=B,A$. $PA+PB=2a=8$. Option (D).