Example 6: Through the right focus $F$ of the ellipse $\frac{x^{2}}{3}+\frac{y^{2}}{2}=1$, draw two perpendicular chords $A B$ and $C D$. Let the midpoints of $A B$ and $C D$ be $M$ and $N$ respectively.
(1) Prove that the line $M N$ must pass through a fixed point, and find this fixed point;
(2) If the slopes of the chords both exist, find the maximum value of the area of $\triangle F M N$.
Now, let's study the general problem:
Through the right focus $F$ of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$, draw two perpendicular chords $A B$ and $C D$. Let the midpoints of $A B$ and $C D$ be $M$ and $N$ respectively.
(1) Prove that the line $M N$ must pass through a fixed point, and find this fixed point;
(2) If the slopes of the chords both exist, find the maximum value of the area of $\triangle F M N$.
(1) Proof: Establish a polar coordinate system with $F$ as the pole and $F x$ as the polar axis. Then the equation of the ellipse is $\rho=\frac{e p}{1+e \cos \theta}$.
Let point $A\left(\rho_{1}, \alpha\right)\left(\rho_{1}=\frac{e p}{1+e \cos \alpha}\right)$,
$B\left(\rho_{2}, \alpha+\pi\right)\left(\rho_{2}=\frac{e p}{1-e \cos \alpha}\right)$,
$C\left(\rho_{3}, \alpha+\frac{\pi}{2}\right)\left(\rho_{3}=\frac{e p}{1-e \sin \alpha}\right)$,
$D\left(\rho_{4}, \alpha+\frac{3 \pi}{2}\right)\left(\rho_{4}=\frac{e p}{1+e \sin \alpha}\right)$,
then point $M\left(\frac{\rho_{1}-\rho_{2}}{2}, \alpha\right), N\left(\frac{\rho_{3}-\rho_{4}}{2}, \alpha+\frac{\pi}{2}\right)$, i.e.,
$M\left(-\frac{e^{2} p \cos \alpha}{1-e^{2} \cos ^{2} \alpha}, \alpha\right), N\left(\frac{e^{2} p \sin \alpha}{1-e^{2} \sin ^{2} \alpha}, \alpha+\frac{\pi}{2}\right)$.
Let $\rho_{5}=-\frac{e^{2} p \cos \alpha}{1-e^{2} \cos ^{2} \alpha}, \rho_{6}=\frac{e^{2} p \sin \alpha}{1-e^{2} \sin ^{2} \alpha}$.
Thus, $-\frac{\cos \alpha}{\rho_{5}}+\frac{\sin \alpha}{\rho_{6}}$
$=\frac{1-e^{2} \cos ^{2} \alpha}{e^{2} p}+\frac{1-e^{2} \sin ^{2} \alpha}{e^{2} p}=\frac{2-e^{2}}{e^{2} p}$.
Let the polar coordinates of any point $P$ on $M N$ be $(\rho, \theta)$.
Then, by the two-point form of the polar equation of a line, we have
$\frac{\sin \left[\left(\alpha+\frac{\pi}{2}\right)-\theta\right]}{\rho_{5}}+\frac{\sin (\theta-\alpha)}{\rho_{6}}=\frac{\sin \frac{\pi}{2}}{\rho}$.
Let $\theta=\pi$, we get
$-\frac{\cos \alpha}{\rho_{5}}+\frac{\sin \alpha}{\rho_{6}}=\frac{1}{\rho}$.
Comparing equations (1) and (2), when $\theta=\pi$, $\rho=\frac{e^{2} p}{2-e^{2}}$, i.e., the point
$\left(\frac{e^{2} p}{2-e^{2}}, \pi\right)$ lies on the line $M N$.
Note that, $c=\frac{e^{2} p}{1-e^{2}}$.
Returning to the Cartesian coordinate system, the line $M N$ passes through the fixed point $\left(\frac{e^{2} p}{1-e^{2}}-\frac{e^{2} p}{2-e^{2}}, 0\right)$, i.e., $\left(\frac{e^{2} p}{\left(1-e^{2}\right)\left(2-e^{2}\right)}, 0\right)$.

For example 6, $e^{2}=\frac{1}{3}, p=\frac{b^{2}}{c}=2$, the line $M N$ passes through the fixed point $\left(\frac{3}{5}, 0\right)$.
(2) Solution: Let $S$ be the area of $\triangle P M N$.
Then $S=\frac{1}{2}\left|\rho_{5}\right|\left|\rho_{6}\right|$
$$
\begin{array}{l}
=\frac{1}{2} \cdot \frac{e^{4} p^{2} \sin \alpha \cdot \cos \alpha}{\left(1-e^{2} \sin ^{2} \alpha\right)\left(1-e^{2} \cos ^{2} \alpha\right)} \\
=\frac{1}{4} \cdot \frac{e^{4} p^{2} \sin 2 \alpha}{1-e^{2}+\frac{e^{4}}{4} \sin ^{2} 2 \alpha}=\frac{e^{4} p^{2} \sin 2 \alpha}{4\left(1-e^{2}\right)+e^{4} \sin ^{2} 2 \alpha} \\
=\frac{e^{4} p^{2}}{e^{4} \sin 2 \alpha+\frac{4\left(1-e^{2}\right)}{\sin 2 \alpha}} .
\end{array}
$$
(i) When $\frac{4\left(1-e^{2}\right)}{e^{4}} \leqslant 1$, i.e., $2(\sqrt{2}-1) \leqslant e^{2}1$, i.e., $0<e^{2}<2(\sqrt{2}-1)$

then $e^{4} t+\frac{4\left(1-e^{2}\right)}{t}$ is monotonically decreasing on the interval $(0,1]$, hence $S \leqslant \frac{e^{4} p^{2}}{e^{4}+4\left(1-e^{2}\right)}=\frac{e^{4} p^{2}}{\left(2-e^{2}\right)^{2}}$.
For example 6, $e^{2}=\frac{1}{3}, p=\frac{b^{2}}{c}=2$, the maximum area is $\frac{4}{25}$.