distance of the 'waist' of a caustic in a parabola I just want to compute the distance of the 'waist' shown in the graphic. (Since I can't upload a graphic I'll show  the link below)
Normal families of a parabola
I don't know how to call it bt 'wist'. What I mean is the distance in the middle. I think it should depend on $x$ and some parameter related to the with of the parabola.
note that the lines are drawn only until $$ x= 1$$ but I want $x$ as a parameter; for any $x$.
Any Ideas? ,
Thanks
By the way, I'm not sure if that is called caustic so be free to correct me if I'm wrong.
 A: The normal line to graph of $f(x)=x^2$ at the point $(s,s^2)$ has slope $-\frac1{2s}$ since it is perpendicular to the tangent which has slope $2s$. Thus the equation is
$$
y=-\tfrac1{2s}(x-s)+s^2=-\tfrac1{2s}\cdot x+s^2+\tfrac12
$$
so the intersection of the normal lines at $(s,s^2)$ and $(t,t^2)$ can be found as follows
$$
-\tfrac1{2s}\cdot x+s^2=-\tfrac1{2t}\cdot x+t^2\\
\iff\\
x=-2st(s+t)
$$
When we fix $s$ and vary $t$ we have a quadratic expression $x=-2s\cdot t^2-2s^2\cdot t$ in $t$ that, assuming $s>0$, attains its maximum at
$$
t=-\frac{-2s^2}{2\cdot (-2s)}=-\frac s2
$$
leading to $x=\frac{s^3}2$. If $s<0$ this becomes a minimum instead. By symmetry we see that the "waist" of the normal family for $x\in[-s,s]$ must be $2\cdot\frac{s^3}2=s^3$. So to conclude:
$$
waist=s^3
$$
I do not know how you want to define the "waist" if we allow domains beyond $x\in[-\sqrt2,\sqrt2]$ since then the legs of the parabola reaches the region produced by the normals. That changes the essential nature of the shape.
