Question

A ray emanating from point $A$ $(-3,0)$ incidents at the point $P$ on the surface of the ellipse $16x^2+25y^2=400$ and gets reflected parallel to minor axis of the ellipse and again reflected at point $Q$ on the surface of the ellipse, then area of $\Delta APQ$ is $\frac{a}{b},$ where $\frac{a}{b}$ is in the smallest form. Then the value of $a+b$ is

Answer 1

Ellipse is $\frac{x^2}{25}$ + $\frac{y^2}{16}=1$
$\therefore a=5,b=4$


$\therefore$ Foci $=(\pm 3,0)$
A line passing through one focus, after reflection, the reflected line passes through the other focus.
Since, reflected ray is parallel to minor axis i.e., $y$-axis,
$\therefore P\equiv (3,\frac{16}{5})$ and $Q\equiv (3,\frac{-16}{5})$
Area of $\Delta APQ=\frac{1}{2}\times 6\times \frac{32}{5}=\frac{96}{5}$