11. Let $P$ be a moving point on the parabola $y^{2}=2 x$, and $B, C$ be points on the $y$-axis. The circle $(x-1)^{2}+y^{2}=1$ is the incircle of $\mathrm{V} P B C$. Find the minimum value of the area of $\mathrm{V} P B C$.
11. Analysis: Let the coordinates of $P$ be $(x, y)$, and the points of tangency of the circle on sides $PB$ and $PC$ be $D$ and $E$, respectively. Then, $S_{\Delta} P B C = \frac{1}{2} |B C| x$. Also, $S_{\Delta} P B C = \frac{1}{2} (|B C| + |P B| + |P C|) \cdot 1 = |B C| + |P D|$, and $|P D|^2 = (x-1)^2 + y^2 - 1 = x^2 - 2x + y^2 = x^2$, so $|P D| = x$. Therefore, $S_{\Delta} P B C = |B C| + x$, and also $S_{\Delta} P B C = \frac{1}{2} |B C| x$, which gives $\frac{1}{2} |B C| x = |B C| + x$. We get $|B C| = \frac{2 x}{x-2} (x > 2)$, thus $S_{\triangle P B C} = \frac{2 x}{x-2} + x = \frac{4}{x-2} + x - 2 + 4 \geq 8$, with equality when $x = 4$. Therefore, the minimum value of the area is 8.