Curvature and symmetry about a point

Motivated by one of the proposed exercises today, we observe that the curve $y=x^3$, which is symmetric about the origin of the plane, satisfies that the curvature at the point $-x$ is the opposite in $x$: look in which side lies the graph of $y=x^3$ around $x$. Thus we prove the next result:

Result. If $\alpha:I\rightarrow{\mathbb R}^2$ is a planar curve which is symmetric about  a point $p_0\in\alpha(I)$, then $\kappa(-s)=-\kappa(s)$. Here we are assuming that $\alpha(0)=p$.

After a translation and without loss of generality, we suppose that $p_0=(0,0)$. Say that $\alpha$ is symmetric about the origin means $\alpha(-s)= M\alpha(s)$, where $M$ is the symmetry, that is, $Mx=-x$. Thus $\alpha(-s)=-\alpha(s)$.  Let $\beta(s)=\alpha(-s)$. Then $\kappa_\beta(s)=-\kappa_\alpha(-s)$. On the other hand, the curve $\gamma(s)=-\alpha(s)$ is a direct rigid motion of $\alpha$, so $\kappa_\gamma(s)=\kappa_\alpha(s)$. Since $\kappa_\gamma(s)=\kappa_\beta(s)$, we conclude $$-\kappa_\alpha(-s)=\kappa_\alpha(s).$$
This can be checked for curves that are graphs of $y=f(x)$ where we know that $$\kappa_\alpha(x)=\frac{f''(x)}{(1+f'(x)^2)^{3/2}}.$$ Denote $\alpha$ the curve $y=f(x)$. As $f$ is symmetric about the origin, then $f(-x)=-f(x)$. The curvature of $y=g(x)$, where $g(x)=f(-x)$ is
$$g'(x)=-f'(-x), g''(x)=f''(-x)\Rightarrow \kappa_g(x)=\frac{f''(-x)}{(1+f'(-x)^2)^{3/2}}.$$
The curvature of $y=-h(x)$ is $$h'(x)=-f'(x), h''(x)=-f''(x)\Rightarrow \kappa_h(x)=\frac{-f''(x)}{(1+f'(x)^2)^{3/2}}.$$ Since the curvatures of both graphs coincide, we have $$\kappa_f(-x)=\frac{f''(-x)}{(1+f'(-x)^2)^{3/2}}=\frac{-f''(x)}{(1+f'(x)^2)^{3/2}}=-\kappa_f(x).$$
The converse is also true:

Theorem. If $\alpha:I=(-a,a)\rightarrow{\mathbb R}^2$ is a planar curve  such that $\kappa(-s)=-\kappa(s)$, then the graphic of $\alpha$ is symmetric about  the point $p=\alpha(0)$.

After a translation, we suppose $\alpha(0)=(0,0)$. Define two curves: $\beta(s)=\alpha(-s)$ and $\gamma(s)=-\alpha(s)$. We compute their curvatures. For $\beta$, $\kappa_\beta(s)=-\kappa(-s)$ and for $\gamma$, $\kappa_\gamma(s)=\kappa(s)$ because the symmetry is a direct rigid motion.  By hypothesis, $\kappa_\gamma(s)=\kappa_\beta(s)$ so there is a direct rigid motion $M$ such that $\gamma(s)=M\beta(s)$. Because $\gamma(0)=\alpha(0)=\beta(0)$ and $\gamma'(0)=\beta'(0)$ (check!), then their normal vector also coincide at $s=0$. Then $M$ must be the identity, proving the result.