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 $y=m x, m \neq 0$ then find the maximum area of quadrilateral ACBD.

Answer 1

Image of $A(h, 0)$ in the line mirror $m x-y=0$
$\frac{x-h}{m}=\frac{y-0}{-1}=-2\left(\frac{m h}{m^2+1}\right) $
$x=\frac{h\left(1-m^2\right)}{1+m^2}, y=\frac{2 m h}{m^2+1}$
$\therefore C=\left(\frac{h\left(1- m ^2\right)}{1+ m ^2}, \frac{2 mh }{ m ^2+1}\right)$
Similarly $D=\left(\frac{-h\left(1-m^2\right)}{1+m^2}, \frac{-2 m h}{m^2+1}\right)$
$A C=\sqrt{\left(h-\frac{h\left(1-m^2\right)}{1+m^2}\right)^2+\frac{4 m^2 h^2}{\left(1+m^2\right)^2}}=\sqrt{\frac{4 m^4 h^2}{\left(1+m^2\right)^2}+\frac{4 m^2 h^2}{\left(1+m^2\right)^2}}=\frac{2 m h \sqrt{1+m^2}}{\left(1+m^2\right)}$
$A D=\sqrt{\left(h+\frac{h\left(1-m^2\right)}{1+m^2}\right)^2+\frac{4 m^2 h^2}{\left(1+m^2\right)^2}}=\sqrt{\frac{4 h^2}{\left(1+m^2\right)^2}+\frac{4 m^2 h^2}{\left(1+m^2\right)^2}}=\frac{2 h}{\left(1+m^2\right)} \sqrt{1+m^2}$
$\therefore$ Area $= AC \cdot AD =\frac{4 mh ^2\left(1+ m ^2\right)}{\left(1+ m ^2\right)^2}=\frac{4 h ^2}{ m +\frac{1}{ m }}$
Area $_{\max }=\frac{4 \times 25}{2}=50$