II. (20 points) Let $P$ be a moving point on the line $y=x-2$. Draw tangents from $P$ to the parabola $y=\frac{1}{2} x^{2}$, with points of tangency being $A$ and $B$.
(1) Prove that the line $AB$ passes through a fixed point;
(2) Find the minimum value of the area $S$ of $\triangle PAB$, and the coordinates of point $P$ when this minimum value is achieved.
Let $P\left(x_{0}, x_{0}-2\right) 、 A\left(x_{1}, \frac{1}{2} x_{1}^{2}\right), B\left(x_{2}, \frac{1}{2} x_{2}^{2}\right)$, and the equation of the chord of tangents gives the line $l_{A B}$:
$$
\begin{array}{l}
\frac{\left(x_{0}-2\right)+y}{2}=\frac{1}{2} x_{0} x \\
\Rightarrow y=x_{0}(x-1)+2 .
\end{array}
$$

Thus, the line $A B$ passes through the fixed point $(1,2)$.
(2) The distance from point $P$ to the line $A B$ is
$$
\begin{array}{l}
d=\frac{\left|x_{0}^{2}-2 x_{0}+4\right|}{\sqrt{x_{0}^{2}+1}}=\frac{x_{0}^{2}-2 x_{0}+4}{\sqrt{x_{0}^{2}+1}} . \\
\text { Also, }|A B|=\sqrt{x_{0}^{2}+1}\left|x_{1}-x_{2}\right| \\
=\sqrt{x_{0}^{2}+1} \cdot \sqrt{\left(x_{1}+x_{2}\right)^{2}-4 x_{1} x_{2}} \\
=2 \sqrt{x_{0}^{2}+1} \cdot \sqrt{x_{0}^{2}-2 x_{0}+4},
\end{array}
$$

Then $S=|A B| d=\sqrt{\left(x_{0}^{2}-2 x_{0}+4\right)^{3}}$.
Notice that
$$
x_{0}^{2}-2 x_{0}+4=\left(x_{0}-1\right)^{2}+3 \geqslant 3 \text {, }
$$

with equality holding if and only if $x_{0}=1$.
Thus, $S_{\min }=3 \sqrt{3}$. At this point, the point $P(1,-1)$.