Example 9 (Mathematics for Middle School, Issue 1, 1994 Olympiad Training Question) Given the ellipse $\frac{x^{2}}{2}+y^{2}=1, D A \perp A B, C B \perp A B$ and $D A=3 \sqrt{2}, C B=\sqrt{2}$, a moving point $P$ is on $\overparen{A B}$. Then the minimum value of the area of $\triangle P C D$ is
Solve: Fill in $4-\sqrt{6}$. Reason: It is easy to find the line $C D: x+y-2 \sqrt{2}=$ 0. Let the tangent point $P_{0}\left(x_{0}, y_{0}\right)$, it is easy to know that the tangent line $E F$ through $P_{0}$ parallel to $C D$ has the minimum distance to $C D$. Above the $x$-axis, the equation of the ellipse is
$$
\begin{array}{l}
y=\sqrt{1-\frac{x^{2}}{2}}, \\
y^{\prime}=\frac{1}{2 \sqrt{1-\frac{x^{2}}{2}}}\left(1-\frac{x^{2}}{2}\right)^{\prime}=-\frac{x}{2 \sqrt{1-\frac{x^{2}}{2}}} .
\end{array}
$$

Then $k_{E F}=-\frac{x_{0}}{2 \sqrt{1-\frac{x_{0}^{2}}{2}}}=-1=k_{C P}$.
Thus, $x_{0}=\frac{2}{\sqrt{3}}, y_{0}=\frac{1}{\sqrt{3}}$. By the point-to-line distance formula, the distance from $P_{0}$ to $C D$ is $2-\frac{\sqrt{6}}{2}$, hence the minimum value of $S_{\triangle F C D}$ is $4-\sqrt{6}$.