(2) (20 points) Let $P$ be a moving point on the line $y=x-2$, and draw the tangent lines from point $P$ to the parabola $y=\frac{1}{2} x^{2}$, with the 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 the minimum value is achieved.
(2) (1) Method 1 Let $P\left(x_{0}, x_{0}-2\right) 、 A\left(x_{1}, \frac{1}{2} x_{1}^{2}\right) 、 B\left(x_{2}, \frac{1}{2} x_{2}^{2}\right)$. From $y=\frac{1}{2} x^{2}$, we get $y^{\prime}=x$. Therefore, $k_{P A}=x_{1}, k_{P B}=x_{2}$.
Thus, the equation of the tangent line $P A$ is $y=x_{1}\left(x-x_{1}\right)+\frac{1}{2} x_{1}^{2}$, which simplifies to $y=x_{1} x- \frac{1}{2} x_{1}^{2}$, and the equation of the tangent line $P B$ is $y=x_{2} x-\frac{1}{2} x_{2}^{2}$.

Since point $P\left(x_{0}, x_{0}-2\right)$ lies on both tangents, we have $x_{0}-2=x_{1} x_{0}-\frac{1}{2} x_{1}^{2}$ and $x_{0}-2=x_{2} x_{0}-\frac{1}{2} x_{2}^{2}$.
From this, we get $x_{1}+x_{2}=2 x_{0}$ and $x_{1} x_{2}=2 x_{0}-4$.
Also, $k_{A B}=\frac{\frac{1}{2} x_{1}^{2}-\frac{1}{2} x_{2}^{2}}{x_{1}-x_{2}}=\frac{x_{1}+x_{2}}{2}$, so the equation of line $A B$ is
$$
y=\frac{x_{1}+x_{2}}{2}\left(x-x_{1}\right)+\frac{1}{2} x_{1}^{2},
$$

which simplifies to $y=x_{0}(x-1)+2$.
Thus, line $A B$ passes through the fixed point $(1,2)$.
Method 2 Let $P\left(x_{0}, x_{0}-2\right)$, by the chord of contact equation, the equation of line $A B$ is $\frac{\left(x_{0}-2\right)+y}{2}=\frac{1}{2} x_{0} x$, which simplifies to $y=x_{0}(x-1)+2$.
Thus, line $A B$ passes through the fixed point $(1,2)$.
(2) The distance from point $P$ to line $A B$
$$
\begin{aligned}
d & =\frac{\left|x_{0}^{2}-2 x_{0}+4\right|}{\sqrt{x_{0}^{2}+1}}=\frac{x_{0}^{2}-2 x_{0}+4}{\sqrt{x_{0}^{2}+1}} . \\
\text { Also, }|A B| & =\sqrt{x_{0}^{2}+1}\left|x_{1}-x_{2}\right|=\sqrt{x_{0}^{2}+1} \cdot \sqrt{\left(x_{1}+x_{2}\right)^{2}-4 x_{1} x_{2}} \\
& =2 \sqrt{x_{0}^{2}+1} \cdot \sqrt{x_{0}^{2}-2 x_{0}+4},
\end{aligned}
$$

Therefore, $S=|A B| d=\left(x_{0}^{2}-2 x_{0}+4\right) \sqrt{x_{0}^{2}-2 x_{0}+4}$.
Since $x_{0}^{2}-2 x_{0}+4=\left(x_{0}-1\right)^{2}+3 \geqslant 3$, with equality if and only if $x_{0}=1$.
Hence, $S_{\text {min }}=3 \sqrt{3}$, at which point the coordinates of $P$ are $(1,-1)$.