Getting the minimum radius of curvature of a conic section I am fairly new to this forum and since I am not directly from a mathmetics background I recently ran into a problem I cannot solve. 
What I am trying to do is to intersect a cone at a specific angle and want to receive the minimum radius of curvature. I know how to do it for a cylinder and I also know that I will get either an ellipse, a parabola or a hyperbole for a conic section, but I cannot find a source for either the euqtion of the intersection nor the minimum radius of curvature/curvature itself. 
I found some theoretical proofs for the different intersections, but unfortunately the math behind it was a bit too high for me. 
Is there maybe a short and clear answer to this question (for a stupid engineer as myself)? Or can someone refer me to an other sourve where I could the infromation from? I would be more than happy to hear from you! Thanks in advance!
 A: First of all, the minimum radius of curvature for a conic section occurs at a vertex, and it has the same length as the semi-latus rectum of the conic, that is $b^2/a$ for an ellipse or hyperbola, where $a$ and $b$ are as usual the semiaxes.
The values of $a$ and $b$ depend not only on the inclination of the intersecting plane, but also on the distance of that plane from the vertex of the cone. I'll express them as a function of the distances $m$ and $n$ of the vertices (of the ellipse or hyperbola) from the vertex of the cone. If $u$ is the semi-aperture angle of the cone, we have: 
$$
4a^2=m^2+n^2\mp2mn\cos2u,\quad b^2=mn\sin^2u,
$$
where sign $-$ must be taken for an ellipse and sign $+$ for a hyperbola. You can find the proof (for an ellipse, but that for a hyperbola is analogous) in this answer.
It follows that the minimum radius of curvature is
$$
r_{MIN}={2mn\sin^2u\over\sqrt{m^2+n^2\mp2mn\cos2u}}
$$
For a parabola, you just need to take the limit of the above result for $n\to\infty$:
$$
r_{MIN}=2m\sin^2u.
$$
