Question

Prove that perpendicular drawn from focus upon any tangent of a parabola lies on the tangent at the vertex.

Answer 1

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-

$y-\frac{c^2}{4a}=\frac{c}{2a}(x-c)$

Let the distance d and find its minimum.

$d=\frac{4a\left(a\right)-2c\left(0\right)-c^2+2c^2}{\sqrt{16a^2+4c^2}}$

$=\frac{4a^2+c^2}{\sqrt{16a^2+4c^2}}$

$=\frac{1}{2}\sqrt{4a^2+c^2}$

this distance has its minimum varying values

of c at c = 0 and so d = a