Prove that the foot of the perpendicular from the focus to any tangent of a parabola lies on the tangent to the vertex

Prove that the foot of the perpendicular from the focus to any tangent of a parabola lies on the tangent to the vertex

I've been trying to prove this by plugging in the negative reciprocal of the slope of the tangent at a point $(x, y)$ into a line which passes through that point and the axis of symmetry. Then I plug the value of the focus into the result and solve for $x$. However the slope is undefined for any line parallel to the axis of symmetry.

Solution 1:

Let $F$ be the focus of the parabola, $HG$ its directrix, with vertex $V$ the midpoint of $FH$. From the definition of parabola it follows that $PF=PG$, where $P$ is any point on the parabola and $G$ its projection on the directrix.

The tangent at $P$ is the angle bisector of $\angle FPG$, hence it is perpendicular to the base $GF$ of isosceles triangle $PFG$, and intersects it at its midpoint $M$.

But the tangent at $V$ is parallel to the directrix and bisects $FH$, hence it also bisects $FG$ at $M$, as it was to be proved.