Question

Let $A$ and $B$ be two points on the major axis of the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1,$ which are equidistant from the centre. If $C$ and $D$ are the images of these points in the line mirror $y=mx(m\ne 0),$ then the maximum area of quadrilateral $ACBD$ is

Answer 1

Image of $A(h,0)$ in the line mirror $mx-y=0:$
$\frac{x-h}{m}=\frac{y-0}{-1}=-2\left(\frac{mh}{m^2+1}\right)$
$\Rightarrow x=\frac{h(1-m^2)}{1+m^2},y=\frac{2mh}{1+m^2}$
$\therefore C\equiv (\frac{h(1-m^2)}{1+m^2},\frac{2mh}{1+m^2})$

Similarly, $D\equiv (\frac{-h(1-m^2)}{1+m^2},\frac{-2mh}{1+m^2})$

$AC=\sqrt{{(h-\frac{h(1-m^2)}{1+m^2})}^2+\frac{4h^2{m}^2}{(1+m^2{)}^2}}=\frac{2mh}{1+m^2}\sqrt{1+m^2}$

$AD=\sqrt{{(h+\frac{h(1-m^2)}{1+m^2})}^2+\frac{4h^2{m}^2}{(1+m^2{)}^2}}=\frac{2h}{1+m^2}\sqrt{1+m^2}$

Area $=AC\cdot AD=\frac{4m{h}^2(1+m^2)}{(1+m^2{)}^2}=\frac{4h^2}{m+\frac{1}{m}}$
Maximum area $=\frac{4\times 25}{2}=50$