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.

Without loss of generality, we consider only the case $x^2=4ay$. the focus is $(0,a)$ and the slope at any point $(c,\frac{c^2}{4a})$ is $\frac{c}{2a}$ and the tangent equation is $$y-\frac{c^2}{4a}=\frac{c}{2a}(x-c)$$ Now you have to get the distance $d$ and find its minimum. 
$$d=\frac{4a(a)-2c(0)-c^2+2c^2}{\sqrt{16a^2+4c^2}}
\\=\frac{4a^2+c^2}{\sqrt{16a^2+4c^2}}
\\=\frac{1}{2}\sqrt{4a^2+c^2}$$ this distace has its minimum varying values of $c$ at $c=0$ and so $d=a$
I hope this helps
