This might be a trivial question, but I can't seem to figure it out. Suppose I have an implicitly defined curve in the plane given by $f(x,y) = t$.


This curve is strictly convex, and feel free to assume as much regularity on $f$ as you'd like. I know I can compute the curvature at a point on this curve via 
$$
\kappa = \frac{|n\_{\perp}^THn\_\perp|}{|\nabla f|}
$$
where $H = \nabla^2 f$ is the Hessian of $f$ and $n\_\perp$ is the unit tangent to the curve. 


**My question is this:** is there any relationship between points of extreme curvature (vertices) and points where the unit tangent vector is actually an eigenvector of the Hessian?


This seems to be true for functions of the form $x^TAx$ but I can't seem to prove it in general.



Vertices of Curves and Eigenvectors of Hessian
Let $\Gamma=\{p|f(p)=0\}$ be smooth. Consider only a small neighbourhood $U$ of some point on $\Gamma$.


Trick: we can replace $f$ by the signed distance function ($f(p)={\rm dist}(p,\Gamma)$ on one side of $\Gamma$, $f(p)=-{\rm dist}(p,\Gamma)$ on the other side). The new $f$ is still smooth and defines the same smooth $\Gamma$, but has an additional property: it satisfies the eikonal equation
$$ |\nabla f|^2 = 1 $$
everywhere in $U$. Take the gradient:
$$ H \nabla f = 0 $$
in $U$. That means $\nabla f$ is an eigenvector of $H$. But $H$ is symmetric, and in any point on $\Gamma$ the unit tangent $n\_\perp$ is orthogonal to $\nabla f$, so it is an eigenvector too.


So the desired relationship does not exist.


