Noted that ellipse properties $d_1+d_2=2D$, focal length, $f=c$ and radius of minor axis $=r$.
Let $d_1=a;d_2=b$
If $ab$ is not maximum, then $\sqrt{ab}$ not maximum
W.l.o.g, prove max $ab=D^2$
$ab=\sqrt{(r-dr)^2+(c-dc)^2}×\sqrt{(r-dr)^2+(c+dc)^2}$
$ab=\sqrt{(r-dr)^4+(r-dr)^2(2c^2+2d^2c)+(c^2-d^2c)^2}$
the expression should be all in the form x -dx, so how do I convert $2c^2+2d^2c$ into x-dx form
