Question

If $P(4,\frac{12}{5})$ is a point on the ellipse such that the normal at it meets the major axis of ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$ at $Q$, and $S,S^{\prime }$ are the focii of the given ellipse, where $S^{\prime }$ lies on positive x-axis, then the ratio $S^{\prime }Q:SQ$ is $\frac{\sqrt{a}}{b}$. The value of $a+b$ is

Answer 1

$e^2=1-\frac{16}{25}=\frac{9}{25}$
$e=\frac{3}{5}$
$\Rightarrow S^{\prime }\equiv (-3,0)$ and $S\equiv (3,0)$
The normal at the point $P$ is the bisector of the angle between $S^{\prime }P$ and $SP$.
Hence, $\frac{S^{\prime }Q}{SQ}=\frac{S^{\prime }P}{SP}=\frac{\sqrt{49+\frac{144}{25}}}{\sqrt{1+\frac{144}{25}}}$
$=\frac{\sqrt{1225+144}}{13}=\frac{\sqrt{1369}}{13}$
$a=1369,b=13$
$a+b=1382$