Example 22 Given the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$, one pair of opposite sides of an inscribed parallelogram pass through its two foci. Find the maximum value of the area of this parallelogram.
Solve: As shown in the figure, with the left focus $F_{1}$ as the pole and the positive direction of the $x$-axis as the polar axis, establish a polar coordinate system. Draw $F_{2} H \perp A B$ through $F_{2}$, with the foot of the perpendicular being $H$. Then the equation of the ellipse is $\rho=\frac{e p}{1-e \cos \theta}$, where $e=\frac{c}{a}$ is the eccentricity, and $p=\frac{a^{2}}{c}-c=\frac{b^{2}}{c}$ is the distance from the left focus to the left directrix. Let the inclination angle of the chord $A B$ passing through $F_{1}$ be $\theta(01$, set $00$, i.e., $f\left(t_{1}\right)>f\left(t_{2}\right)$, so $f(t)$ is strictly monotonically decreasing on $(0,1]$, hence $f(t) \geqslant f(1)=b^{2}+c^{2}=a^{2}$, and $S_{\square \triangle A B C D} \leqslant \frac{4 a b^{2} c}{a^{2}}=\frac{4 b^{2} c}{a}$ (equality holds if and only if $t=1$), i.e., the maximum value of $S_{C 7 \text{ CAECD }}$ is $\frac{4 b^{2} c}{a}$.

In summary, when $01$, the maximum value of $S_{\triangle \triangle A B C D}$ is $\frac{4 b^{2} c}{a}$.