jmc

The focus of the parabola $y^2=4ax$ is $F=(a,0),$ and the directrix is $x=-a.$ Let $F^{\prime },$ $P^{\prime },$ $Q^{\prime },$ and $R^{\prime }$ be the projections of $F,$ $P,$ $Q,$ and $R$ onto the directrix, respectively. Let $p=P{P}^{\prime }=PF,$ $q=Q{Q}^{\prime }=QF,$ $a=P^{\prime }{F}^{\prime },$ and $B=Q^{\prime }{F}^{\prime }.$ Since $P,$ $F,$ and $Q$ are collinear, $$\frac{p}{q}=\frac{a}{b}.$$

Then $$\frac{F^{\prime }{Q}^{\prime }}{F^{\prime }{R}^{\prime }}=\frac{b}{a}=\frac{q}{p}=\frac{Q{Q}^{\prime }}{R{R}^{\prime }}.$$This means triangles $F^{\prime }{Q}^{\prime }Q$ and $F^{\prime }{R}^{\prime }R$ are similar, so line $QR$ intersects the $x$-axis at $F^{\prime }=\boxed{(-a,0)}.$