Example 1 As shown in Figure $1, P$ 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 $\triangle P B C$.
(2008, National High School Mathematics Competition)
Explanation: By the symmetry of the parabola, we can assume
$$
P\left(2 t^{2}, 2 t\right)(t>0) \text {. }
$$

Since the equation of the circle is $x^{2}+y^{2}-2 x=0$, the equation of the chord of contact $M N$ is
$$
2 t^{2} x+2 t y-\left(x+2 t^{2}\right)=0,
$$

which simplifies to $\left(2 t^{2}-1\right) x+2 t y-2 t^{2}=0$.
Thus, the double tangents $P B$ and $P C$ are the quadratic curve passing through all common points of the double line
$$
\left[\left(2 t^{2}-1\right) x+2 t y-2 t^{2}\right]^{2}=0
$$

and the circle, with the equation
$$
(x-1)^{2}+y^{2}-1+\lambda\left[\left(2 t^{2}-1\right) x+2 t y-2 t^{2}\right]^{2}=0 .
$$

Substituting the coordinates of point $P$ into the above equation gives $\lambda=-\frac{1}{4 t^{4}}$. Therefore, the equation of the double tangents $P B$ and $P C$ can be written as
$$
(x-1)^{2}+y^{2}-1-\frac{1}{4 t^{4}}\left[\left(2 t^{2}-1\right) x+2 t y-2 t^{2}\right]^{2}=0 \text {. }
$$

Setting $x=0$ in the above equation, we get
$$
y=\frac{1}{t} y-1 \text { or } y=1-\frac{1}{t} y \text {. }
$$

Since when $t=1$, there is only one tangent $P B$ intersecting the axis, and when $01$.
Thus, $y_{B}=\frac{t}{1+t}, y_{C}=\frac{t}{1-t}$,
$$
\begin{array}{l}
B C=\frac{t}{1+t}-\frac{t}{1-t}=\frac{2 t^{2}}{t^{2}-1} . \\
\text { Hence } S_{\triangle P B C}=\frac{1}{2} B C\left|x_{P}\right|=\frac{1}{2} \cdot \frac{2 t^{2}}{t^{2}-1} \cdot 2 t^{2} \\
=\frac{2 t^{4}}{t^{2}-1}=2\left[2+\left(t^{2}-1\right)+\frac{1}{t^{2}-1}\right] \geqslant 8 .
\end{array}
$$

The equality holds if and only if $t=\sqrt{2}$. Therefore, the minimum value sought is 8.