14. Let $0<a<b$, draw lines $l$ and $m$ through the fixed points $A(a, 0)$ and $B(b, 0)$ respectively, such that they intersect the parabola $y^{2}=x$ at four distinct points. When these four points are concyclic, find the locus of the intersection point $P$ of lines $l$ and $m$.
Let $l: x=s y+a, m: x=t y+b$, then the equation of the curve system passing through the four intersection points can be expressed as
$$
(s y-x+a)(t y-x+b)+\lambda\left(y^{2}-x\right)=0
$$

When $(*)$ represents a circle, the equation has no $x y$ term, so $s+t=0$.
Solving $\left\{\begin{array}{l}x=s y+a, \\ x=t y+b\end{array} \Rightarrow x=\frac{a+b}{2}, y=\frac{b-a}{2 s}\right.$.
Conversely, when the lines $l: x=s y+a, m: x=-s y+b$ intersect at point $P\left(\frac{a+b}{2}, \frac{b-a}{2 s}\right)$,
Solving $\left\{\begin{array}{l}x=s y+a, \\ y^{2}=x\end{array} \Rightarrow y^{2}-s y-a=0\right.$
$$
\begin{array}{l}
\Rightarrow|P C| \cdot|P D|=\left(1+s^{2}\right)\left|\left(\frac{b-a}{2 s}-y_{C}\right)\left(\frac{b-a}{2 s}-y_{D}\right)\right| \\
=\left(1+s^{2}\right)\left|\left(\frac{b-a}{2 s}\right)^{2}-\frac{b-a}{2 s}\left(y_{C}+y_{D}\right)+y_{C} y_{D}\right|=\left(1+s^{2}\right)\left|\left(\frac{b-a}{2 s}\right)^{2}-\frac{a+b}{2}\right| .
\end{array}
$$

Similarly, $|P E| \cdot|P F|=\left(1+s^{2}\right)\left|\left(\frac{b-a}{2 s}\right)^{2}-\frac{a+b}{2}\right|=|P C| \cdot|P D|$.
Thus, points $C, D, E, F$ are concyclic, so the locus of intersection point $P$ is the line $x=\frac{a+b}{2}$ excluding the three intersection points with the parabola $y^{2}=x$ and the $x$-axis.