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Given a point on the circumference of a cylinder of radius $r$. The radius of curvature of the circle obtained by cutting the cylinder by a plane perpendicular to central axis and passing through this fixed point is $1/r$ (as we get a circle).

Now suppose the cutting plane is tilted by an angle $\theta$ with respect to the cylinders central axis, then we obtain an ellipse. What will be the curvature of this ellipse in terms of $r$ and $\theta$?

Would it(radius) be $\frac{r}{\sin(\theta)}$?

How do we go about solving this problem?

Assume that the point of interest for curvature is the point touching the minor axis of the ellipse.

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    Ellipses don't have constant curvature; so your expression will need to depend on the position on the ellipse. – 2017-02-03
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    The radius of curvature will change depending on whether you are evaluating the pointy end of the ellipse, or the fat end of the ellipse. However, at the the curvature is proportional to the radius of the largest circle that kisses the curve. And at the pointy end, that is still the radius of the cylinder. – 2017-02-03
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    See here: https://www.encyclopediaofmath.org/index.php/Meusnier_theorem – 2017-02-03

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