Largest bifocal triangle in an ellipse Ellipse $E$ has foci at $P$ and $Q$, and semi-major and semi-minor axes of length $a$ and $b$, respectively.
Find the area of the largest triangle that can be (parttially) inscribed in ellipse $E$, such that one of the vertices is at $P$, the opposite side passes through $Q$, and the two vertices not at $P$ are on the ellipse.
Such triangles have nice properties when the ellipse is considered as a curved reflective surface; for example, if you sent a ray of light from $P$ along one leg of the triangle, the path of the light will be a "figure 8" starting from that triangle, then tracing a triangle on the other side of $P$, and possibly, when the area is maximal, repeating itself.  But I have not been able to prove this.
Extra kudos if somebody can solve this by some geometric argument, without brute-forcing by analytic geometry and calculus.
 A: Using polar coordinates relative to $Q$, $$r(\theta)=\frac{a(1-e^2)}{1-e\cos \theta}=\frac{b^2}{a-c\cos \theta}$$
where $\, \displaystyle c=ae=\frac{1}{2} PQ \,$ and 
$\, \displaystyle e=\sqrt{1-\frac{b^2}{a^2}}$.
For $\, 0< \theta <\pi \,$, we have
\begin{align*}
  \Delta &=
   \frac{PQ\sin \theta}{2}
    \left[
      \frac{a(1-e^2)}{1-e\cos \theta}+\frac{a(1-e^2)}{1+e\cos \theta}
    \right] \\
  &= \frac{2a^2 e(1-e^2)\sin \theta}{1-e^2 \cos^2 \theta} \\
  \frac{d\Delta}{d\theta} &=
  \frac{2a^2e(1-e^2)\cos \theta (e^2 \sin^2 \theta+e^2-1)}
       {(1-e^2 \cos^2 \theta)^2} \\
\end{align*}
Note that $\displaystyle \frac{d\Delta}{d\theta}=0 \,$ when 
$\, \cos \theta =0 \,$ or 
$\displaystyle \, \sin \theta=\frac{\sqrt{1-e^2}}{e} \, , \;
e> \sqrt{\frac{1}{2}}$.
The maximal area occurs when
$$\theta=
\left \{
  \begin{array}{ccl}
    \frac{\pi}{2} & , & 0<e \leq \sqrt{\frac{1}{2}} \\
    \sin^{-1} \frac{\sqrt{1-e^2}}{e} &, & \sqrt{\frac{1}{2}}<e<1
   \end{array}
\right.$$
that is
$$\Delta=
\left \{
  \begin{array}{ccl}
    \frac{2bc^2}{a} & , & 0<e \leq \sqrt{\frac{1}{2}} \\
    ab &, & \sqrt{\frac{1}{2}}<e<1
   \end{array}
\right.$$
A: Well, well, once again I lose the scoop by not master English prose. I give you below my answer before it is finish.
A straight line passing through the foci Q, cuts the ellipse in two points $(x_1,y_1)$ and $(x_2,y_2)$. One realizes that the problem reduces just to determining the maximum of the sum
$y_1+y_2$ (This sum is easily get from the coefficient of $y$ in the respective quadratic resolvent in $y$) because all of the concerned triangles have an area which is the sum of the areas of two triangles having a constant base. 
This base has a length equal to the length of the segment $\overline{PQ}=2c$ where $(c,0)$ is the foci $Q$ to the right of $P$. 
One has $$c^2=a^2-b^2\iff \frac{c^2}{a^2}+\frac{(\frac{b^2}{a})^2}{b^2}= 1$$ It follows  the point $(c,y)\in E$ corresponds to the two coordinates $y=\pm \frac{b^2}{a}$ in the ellipse $E$ of equation $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$. Hence the triangle of vertices $(-c,0),(c,\frac{b^2}{a}),(c,-\frac{b^2}{a})$ has area $$A=\frac{2c\cdot 2b^2}{2a}=\frac{2cb^2}{a}$$ This triangle is a main candidate to be the required maximal one. I stop here.  I suppose the answer of the friend Ng Chung Tak is correct.

