Example 1. (1993, National High School Mathematics Competition) Let $0<a<b$, and draw lines $l$ and $m$ through the fixed points $A(a, 0)$ and $B(b, 0)$, respectively, intersecting the parabola $y^{2}=x$ at four distinct points. When these four points are concyclic, find the locus of the intersection point $P$ of $l$ and $m$.
Let $l$ and $m$ intersect at $P\left(x^{\prime}, y^{\prime}\right)$, then the equations of $l$ and $m$ are:
$$
\begin{array}{l}
y^{\prime} x-\left(x^{\prime}-a\right) y+a y^{\prime}=0, \\
y^{\prime} x-\left(x^{\prime}-b\right) y+b y^{\prime}=0 .
\end{array}
$$

And $F: y^{2}-x=0$.
By Theorem 2, we get
$$
y^{\prime} \cdot\left[2 x^{\prime}+(-a-b)\right]=0 .
$$

Therefore, $y^{\prime}=0$ or $x^{\prime}=\frac{a+b}{2}$.
Verification: $y^{\prime}=0$ is the $x$-axis, which does not meet the problem's requirements. $x^{\prime}=\frac{a+b}{2}$ intersects the parabola $y^{2}=x$ at two points, which also do not meet the problem's requirements. Therefore, the locus of point $P$ is the perpendicular bisector of segment $AB$, but excluding the points where it intersects the $x$-axis and the parabola.