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.
Consider the general ellipse:
$$
\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0) \text {. }
$$

Using $F_{1}$ as the pole and $F_{1} x$ as the polar axis to establish a polar coordinate system, the equation of the ellipse becomes:
$$
\rho=\frac{e p}{1+e \cos \theta}=\frac{b^{2}}{a-c \cos \theta} .
$$

The length of the chord passing through point $F_{1}$ is:
$$
A B=\frac{b^{2}}{a-c \cos \theta}+\frac{b^{2}}{a+c \cos \theta}=\frac{2 a b^{2}}{a^{2}-c^{2} \cos ^{2} \theta} \text {. }
$$

Thus, the area of $\square A B C D$ is:
$$
\begin{array}{l}
S=A B \cdot F_{1} F_{2} \sin \theta \\
=\frac{4 a b^{2} c \sin \theta}{a^{2}-c^{2} \cos ^{2} \theta}=\frac{4 a b^{2} c}{c^{2} \sin \theta+\frac{b^{2}}{\sin \theta}} .
\end{array}
$$

Let $f(\theta)=c^{2} \sin \theta+\frac{b^{2}}{\sin \theta}\left(0<\theta<\pi\right)$. Then, $f'(\theta)=c^{2} \cos \theta-\frac{b^{2} \cos \theta}{\sin ^{2} \theta}=\cos \theta\left(c^{2}-\frac{b^{2}}{\sin ^{2} \theta}\right)$. 

When $0<\theta<\frac{\pi}{2}$, $\cos \theta>0$. Thus, $f'(\theta)>0$ if $\sin \theta>\frac{b}{c}$, and $f'(\theta)<0$ if $\sin \theta<\frac{b}{c}$. Therefore, $f(\theta)$ is decreasing on the interval $\left(0, \frac{b}{c}\right]$ and increasing on the interval $\left[\frac{b}{c}, \frac{\pi}{2}\right)$. 

When $\frac{\pi}{2}<\theta<\pi$, $\cos \theta<0$. Thus, $f'(\theta)<0$ if $\sin \theta>\frac{b}{c}$, and $f'(\theta)>0$ if $\sin \theta<\frac{b}{c}$. Therefore, $f(\theta)$ is increasing on the interval $\left(\frac{\pi}{2}, \frac{b}{c}\right]$ and decreasing on the interval $\left[\frac{b}{c}, \pi\right)$. 

Hence, the minimum value of $f(\theta)$ is $f\left(\frac{b}{c}\right)=2 b c$.

Let $g(t)=c^{2} t+\frac{b^{2}}{t}\left(0<t<1\right)$. Then, $g'(t)=c^{2}-\frac{b^{2}}{t^{2}}$. 

When $0<t<\frac{b}{c}$, $g'(t)<0$, and when $t>\frac{b}{c}$, $g'(t)>0$. Therefore, $g(t)$ is decreasing on the interval $\left(0, \frac{b}{c}\right]$ and increasing on the interval $\left[\frac{b}{c}, 1\right)$. 

Thus, 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, $a=2, b=\sqrt{3}, c=1$, so $S$ reaches its maximum value 6.