Let points $P$ and $Q$ be the intersections of the parabola $y^2 = 4ax,$ where $a > 0,$ and an arbitrary line passing through the focus of the parabola.  Let $R$ be the reflection of $P$ in the $x$-axis.  Find the intersection of line $QR$ with the $x$-axis.
The focus of the parabola $y^2 = 4ax$ is $F = (a,0),$ and the directrix is $x = -a.$  Let $F',$ $P',$ $Q',$ and $R'$ be the projections of $F,$ $P,$ $Q,$ and $R$ onto the directrix, respectively.  Let $p = PP' = PF,$ $q = QQ' = QF,$ $a = P'F',$ and $B = Q'F'.$  Since $P,$ $F,$ and $Q$ are collinear,
\[\frac{p}{q} = \frac{a}{b}.\][asy]
unitsize(1 cm);

real y;
pair F, P, Q, R, S;
pair Fp, Pp, Qp, Rp;

F = (1,0);

path parab = ((-4)^2/4,-4);

for (y = -4; y <= 4; y = y + 0.01) {
  parab = parab--(y^2/4,y);
}

P = intersectionpoint(F--(F + 5*(1,2)),parab);
Q = intersectionpoint(F--(F - 5*(1,2)),parab);
R = reflect((0,0),(1,0))*(P);
S = extension(Q,R,(0,0),(1,0));
Fp = (-1,0);
Pp = (-1,P.y);
Qp = (-1,Q.y);
Rp = (-1,R.y);

draw(parab,red);
draw(P--Q);
draw(P--R);
draw(S--R);
draw((-2,0)--(4,0));
draw((0,-4)--(0,4));
draw((-1,-4)--(-1,4),dashed);
draw(P--Pp);
draw(Q--Qp);
draw(R--Rp);

label("$x = -a$", (-1,-4), dir(270));
label("$p$", (P + Pp)/2, N, red);
label("$p$", (P + F)/2, SE, red);
label("$q$", (Q + Qp)/2, dir(270), red);
label("$q$", (Q + F)/2, SE, red);
label("$a$", (Pp + Fp)/2, W, red);
label("$b$", (Qp + Fp)/2, W, red);
label("$p$", (Rp + R)/2, dir(270), red);

dot("$F$", F, SE);
dot("$P$", P, N);
dot("$Q$", Q, dir(270));
dot("$R$", R, dir(270));
dot("$F'$", S, NW);
dot("$P'$", Pp, W);
dot("$Q'$", Qp, W);
dot("$R'$", Rp, W);
[/asy]

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