Question:
If the co-ordinates of two points $\mathrm{A}$ and $\mathrm{B}$ are $(\sqrt{7}, 0)$ and $(-\sqrt{7}, 0)$ respectively and $P$ is any point on the conic, $9 x^{2}+16 y^{2}=144$, then $\mathrm{PA}+\mathrm{PB}$ is equal to :
Correct Option: , 2
Solution:
Ellipse : $\frac{x^{2}}{16}+\frac{y^{2}}{9}=1$,
$a=4, b=3, c=\sqrt{16-9}=\sqrt{7}$
$\therefore(\pm \sqrt{7}, 0)$ are the foci of given ellipse. So for any point
$P$ on it; $P A+P B=2 a$
$\Rightarrow P A+P B=2(4)=8$
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