# 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.

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.

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

If you were to use a standard parabola like $$y^2=4ax$$, the usual way of representing a point on the parabola is via parametric equations $$x=at^2$$ and $$y=2at$$, so the general point is $$(at^2, 2at)$$.

The gradient of the tangent line to this point is thus $$\frac{d(2at)} {dt}$$ divided by $$\frac{d(at^2)} {dt}$$ i.e. $$\frac{1}{t}$$.

Thus the equation of the tangent line is $$y=\frac{x} {t} + constant$$ or $$constant = ty - x$$. We know the tangent line passes through $$(at^2,2at)$$, so substituting these values for $$x$$ and $$y$$ we get $$constant= at^2$$ and so the equation for our tangent is $$yt - x = at^2$$

The perpendicular through the focus must thus have gradient $$-t$$ and we know it passes through $$(a,0)$$. The equation of this line can be written $$constant=y+tx$$. Substituting $$(a,0)$$ for $$(x,y)$$ in this equation gives $$constant=at$$. Thus $$y+tx=at$$ is the equation of the perpendicular to the tangent through the focus.

Multiply both sides of this last equation by $$t$$ in order to eliminate terms in $$y$$ by subtracting the first equation to get $$t^2x+x=0$$, which can only be true if $$x=0$$. For $$y^2=4ax$$, $$x=0$$ is the equation of the vertex.