Question 168: In the Cartesian coordinate system $x 0 y$, point $P$ is a moving point on the line $y=-x-2$. A tangent line is drawn from point $P$ to the parabola $y=\frac{x^{2}}{2}$, with the points of tangency being $A$ and $B$. Then the minimum value of the area of $\triangle P A B$ is $\qquad$ -
Question 168, Solution: Let $P\left(x_{0}, y_{0}\right)$, then the equation of line $AB$ is $\frac{y+y_{0}}{2}=\frac{x_{0} x}{2}$. Noting that $y_{0}=-x_{0}-2$, the equation of line $AB$ becomes: $x_{0} x-y+x_{0}+2=0$.

Solving the system of equations $\left\{\begin{array}{c}x_{0} x-y+x_{0}+2=0 \\ y=\frac{x^{2}}{2}\end{array}\right.$, we get $x^{2}-2 x_{0} x-2\left(x_{0}+2\right)=0$. By Vieta's formulas, we have: $\left\{\begin{array}{c}x_{A}+x_{B}=2 x_{0} \\ x_{A} x_{B}=-2\left(x_{0}+2\right),\end{array}\right.$ Therefore,
$$
|\mathrm{AB}|=\sqrt{1+\mathrm{x}_{0}^{2}}\left|\mathrm{x}_{\mathrm{A}}-\mathrm{x}_{\mathrm{B}}\right|=\sqrt{1+\mathrm{x}_{0}^{2}} \cdot \sqrt{4 \mathrm{x}_{0}^{2}+8\left(\mathrm{x}_{0}+2\right)}=2 \sqrt{1+\mathrm{x}_{0}^{2}} \cdot \sqrt{\mathrm{x}_{0}^{2}+2 \mathrm{x}_{0}+4}
$$

The distance from $P\left(x_{0},-x_{0}-2\right)$ to line $AB$ is $\frac{\left|x_{0}^{2}-\left(-x_{0}-2\right)+x_{0}+2\right|}{\sqrt{1+x_{0} 2}}=\frac{\left|x_{0}^{2}+2 x_{0}+4\right|}{\sqrt{1+x_{0}^{2}}}$, hence
$$
\begin{array}{l}
S_{\triangle A B C}=\frac{1}{2} \cdot 2 \sqrt{1+x_{0}^{2}} \cdot \sqrt{x_{0}^{2}+2 x_{0}+4} \cdot \frac{\left|x_{0}^{2}+2 x_{0}+4\right|}{\sqrt{1+x_{0}^{2}}} \\
=\left(\left(x_{0}+1\right)^{2}+3\right)^{\frac{3}{2}} \geq 3 \sqrt{3}
\end{array}
$$

Thus, the minimum area of $\triangle PAB$ is $3 \sqrt{3}$, which occurs when $P(-1,-1)$.