# Parallel Curves to a Parabola

I have been modelling parallel curves to a parabola and realise if the parallel curve to a parabola is offset enough then the curve will overlap.

I came across this research paper to explain why a parallel curve to a parabola is not a parabola: http://apollonius.math.nthu.edu.tw/d1/disk5/js/geometry/osc/0.pdf However it does not give a mathematical explanation as to why the parabola invokes or evokes (in laymans terms overlaps itself)

Does it have anything to go with the focal length. ie If the focal length is less than the curve offset then it will invoke/evoke

Any I using the terms evoke and invoke correctly to state that a parallel curve overlaps itself.

We can parametrize the parabola by $c(t)=(t,\frac{t^2}{4r sin(\theta)})$. In finding the parallel curves, we take the normal vector at t, $n(t)=R^{90}[c'(t)]$ where $R^{90}$ is rotation by $90^{\circ}$, and compute $l_s(t)=c(t)+s\space n(t)$ for each s. Now, fix $t=t_0$. I claim there exist $s$ such that $c(t_0)+s\space n(t_0)=c(t_1)+s\space n(t_1)$. This is a sufficient condition of a parallel curve intersecting itself. To simply calculations, I will just compute for $c(t)=(t,t^2)$, but you can see that this is true for any parabola. $c'(t)=(1,2t)$ so that $n(t)=(-2t,1).$ So we want $(t_0,t_0^2)+s (-2t_0,1)=(t_1,t_1^2)+s (-2t_1,1)$, that is, $(t_0-2t_0\space s,t_0^2+s)=(t_1-2t_1\space s,t_1^2+s)$. By inspection, we see this could be true only if $t_0=-t_1$ so that $t_0-2t_0\space s=2t_0\space s-t_0$ which implies $s=1/2$ provided that $t_0,t_1\neq0$. Hence, we have a self-intersection point $(t_0-\frac{1}{2}2t_0,t_0^2+\frac{1}{2})=(0,t_0^2+\frac{1}{2})$.