(15) As shown in the figure, $P$ is a moving point on the parabola $y^{2}=2 x$, points $B$ and $C$ are on the $y$-axis, and the circle $(x-1)^{2}+y^{2}=1$ is inscribed in $\triangle P B C$. Find the minimum value of the area of $\triangle P B C$.
15 Let $P\left(x_{0}, y_{0}\right) 、 B(0, b) 、 C(0, c)$, and assume $b>c$. The equation of the line $P B$ is
$$
y-b=\frac{y_{0}-b}{x_{0}} x
$$

Simplifying, we get
$$
\left(y_{0}-b\right) x-x_{0} y+x_{0} b=0.
$$

Since the distance from the center $(1,0)$ to $P B$ is 1, we have
$$
\frac{\left|y_{0}-b+x_{0} b\right|}{\sqrt{\left(y_{0}-b\right)^{2}+x_{0}^{2}}}=1
$$

Thus,
$$
\left(y_{0}-b\right)^{2}+x_{0}^{2}=\left(y_{0}-b\right)^{2}+2 x_{0} b\left(y_{0}-b\right)+x_{0}^{2} b^{2},
$$

Given that $x_{0}>2$, the above equation simplifies to
$$
\left(x_{0}-2\right) b^{2}+2 y_{0} b-x_{0}=0,
$$

Similarly,
$$
\left(x_{0}-2\right) c^{2}+2 y_{0} c-x_{0}=0.
$$

Therefore,
$$
b+c=\frac{-2 y_{0}}{x_{0}-2}, \quad b c=\frac{-x_{0}}{x_{0}-2},
$$

Then,
$$
(b-c)^{2}=\frac{4 x_{0}^{2}+4 y_{0}^{2}-8 x_{0}}{\left(x_{0}-2\right)^{2}}
$$

Since $P\left(x_{0}, y_{0}\right)$ is a point on the parabola, we have $y_{0}^{2}=2 x_{0}$, thus
$$
(b-c)^{2}=\frac{4 x_{0}^{2}}{\left(x_{0}-2\right)^{2}},
$$

which gives
$$
b-c=\frac{2 x_{0}}{x_{0}-2}
$$

Therefore,
$$
\begin{aligned}
S_{\triangle P B C} & =\frac{1}{2}(b-c) \cdot x_{0} \\
& =\frac{x_{0}}{x_{0}-2} \cdot x_{0} \\
& =\left(x_{0}-2\right)+\frac{4}{x_{0}-2}+4 \\
& \geqslant 2 \sqrt{4}+4 \\
& =8,
\end{aligned}
$$

Equality holds when $\left(x_{0}-2\right)^{2}=4$, i.e., $x_{0}=4, y_{0}= \pm 2 \sqrt{2}$. Therefore, the minimum value of $S_{\triangle P B C}$ is 8.