Example 1. Let $0<a<b$, draw lines $l$ and $m$ through the fixed points $A(a, 0), B(b, 0)$ respectively, intersecting the parabola $y^{2}=x$ at two distinct points. When these four points are concyclic, find the locus of the intersection point $P$ of $l$ and $m$.
Let $P\left(x^{\prime}, y^{\prime}\right)$, then the equations of $l$ and $m$ are:
$$
\begin{array}{l}
\quad l: y^{\prime} x-\left(x^{\prime}-a\right) y \\
+a y^{\prime}=0, \\
\quad m: y^{\prime} x-\left(x^{\prime}-b\right) y \\
+b y^{\prime}=0 .
\end{array}
$$

And $f: y^{2}-x=0$, so by Theorem 2 we get
$$
y^{\prime}\left[2 x^{\prime}+(\cdots a-b)\right]=0 \text{. }
$$

Thus, $y^{\prime}=0$ (which is not valid for $x$) or $x^{\prime}=\frac{a+b}{2}$ (the perpendicular bisector of AB, but excluding its intersection points with the parabola and the $x$-axis)