14. Given point $A(a, b)$, parabola
$$
C: y^{2}=2 p x(a \neq 0, b \neq 0, a \neq 2 p) \text {. }
$$

Draw a line $l$ through point $A$, intersecting parabola $C$ at points $P$ and $Q$. If the circle with diameter $PQ$ passes through the vertex of parabola $C$, find the equation of line $l$.
14. If the line $l$ passes through the origin, it obviously meets the requirement, and the equation is $y=\frac{b}{a} x$.
If the line $l$ does not pass through the origin, let its equation be
$$
x=m(y-b)+a \text {. }
$$

Let $P\left(x_{1}, y_{1}\right) 、 Q\left(x_{2}, y_{2}\right)$, then
$$
O P \perp O Q \Leftrightarrow x_{1} x_{2}+y_{1} y_{2}=0 \text {. }
$$

Since $y_{1}^{2}=2 p x_{1}, y_{2}^{2}=2 p x_{2}$, we have $y_{1} y_{2}=-4 p^{2}$.
From the system of equations $\left\{\begin{array}{l}x=m(y-b)+a, \\ y^{2}=2 p x,\end{array}\right.$ eliminating $x$ gives
$$
y^{2}-2 p m y+2 p(m b-a)=0 \text {. }
$$

By Vieta's formulas, we get
$$
y_{1} y_{2}=2 p(m b-a)=-4 p^{2} \text {. }
$$

Thus, $m=\frac{a-2 p}{b}$.
Therefore, the required equation is
$$
b x-(a-2 p) y-2 b p=0 .
$$

Since $-4 p^{2}<0$, we have $-2 p(a-b m)<0$, which means the constant term of equation (1) is negative. Hence, the discriminant is greater than 0, and equation (1) definitely has solutions $y_{1} 、 y_{2}$. Therefore, equation (2) meets the requirement.