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 find the maximum area of quadrilateral ACBD.

Answer 1

Answer: 50

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