Show that all center of curvature to the curve = sin²x/x² at its points of contact with the -axis lie on a parabola. How to solve this problem?

Question

Can someone please help me solve this question. I tried to solve it by finding the radius of the curvature but the derivatives are huge. Is there any simple way to do this?

Answer 1

First of all, let observe that in the parabola $y=ax^2$ the center of curvature in the vertex is $r=1/2a.$

Similarly, for the function $$ f(x)=\frac{\sin^2x}{x^2} $$ it is enough to make a second order approximation in the minima, i.e $$ f(x)=\frac{(x-k\pi)^2}{(k\pi)^2}+o(x-k\pi)^2 $$ so the radius in $M_k=(k\pi,0)$ is $r_k=1/2a=(k\pi)^2/2$ and the centers are $$ C_k=\left(k\pi,\frac{(k\pi)^2}{2}\right) $$