Example 5 Given the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$ and an inscribed parallelogram with one pair of opposite sides passing through the foci $F_{1}$ and $F_{2}$ of the ellipse. Find the maximum area of the parallelogram. ${ }^{[4]}$
(2013, National High School Mathematics League Shandong Province Preliminary Contest)
Solve the general ellipse:
$$
\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)
$$

Corresponding problems.
With $F_{1}$ as the pole and $F_{1} x$ as the polar axis, the equation of the ellipse is transformed into
$$
\rho=\frac{e p}{1+e \cos \theta}=\frac{b^{2}}{a-c \cos \theta} .
$$

Therefore, the length of the chord passing through point $F_{1}$
$$
\begin{array}{l}
A B=\frac{b^{2}}{a-c \cos \theta}+\frac{b^{2}}{a+c \cos \theta} \\
=\frac{2 a b^{2}}{a^{2}-c^{2} \cos ^{2} \theta},
\end{array}
$$
The area of $\square A B C D$
$$
\begin{aligned}
S & =A B \cdot F_{1} F_{2} \sin \theta=\frac{4 a b^{2} c \sin \theta}{a^{2}-c^{2} \cos ^{2} \theta} \\
& =\frac{4 a b^{2} c \sin \theta}{b^{2}+c^{2} \sin ^{2} \theta}=\frac{4 a b^{2} c}{c^{2} \sin \theta+\frac{b^{2}}{\sin \theta}} . \\
& \text { Let } f(\theta)=c^{2} \sin \theta+\frac{b^{2}}{\sin \theta}\left(0<\theta<\pi\right) . \text { Then } f^{\prime}(\theta)=c^{2} \cos \theta-\frac{b^{2} \cos \theta}{\sin ^{2} \theta}=\frac{\cos \theta\left(c^{2} \sin ^{2} \theta-b^{2}\right)}{\sin ^{2} \theta} .
\end{aligned}
$$
Let $f^{\prime}(\theta)=0$, we get $\sin \theta=\frac{b}{c}$. When $\theta \in\left(0, \frac{b}{c}\right)$, $f^{\prime}(\theta)<0$; when $\theta \in\left(\frac{b}{c}, \pi\right)$, $f^{\prime}(\theta)>0$. Therefore, $f(\theta)$ is monotonically decreasing in the interval $\left(0, \frac{b}{c}\right]$ and monotonically increasing in the interval $\left[\frac{b}{c}, \pi\right)$. Hence,
(1) When $\frac{b}{c} \geqslant 1$, i.e., $b\sqrt{2} b$, $g(t)$ is monotonically decreasing in the interval $\left(0, \frac{b}{c}\right]$ and monotonically increasing in the interval $\left[\frac{b}{c}, 1\right)$. Therefore, the minimum value of $g(t)$ is $g\left(\frac{b}{c}\right)=2 b c$. At this point, $S$ reaches its maximum value $\frac{4 a b^{2} c}{2 b c}=2 a b$.
In summary, when $b\sqrt{2} b$, $S$ reaches its maximum value $2 a b$.
For this problem, $a=2, b=\sqrt{3}, c=1, S$ reaches its maximum value $\frac{4 b^{2} c}{a}=6$.