Suppose you have a solid ellipse with axes $a$ and $b$, $(x/a)^2 + (y/b)^2 = 1$, confined inside a unit-radius circle. You shake the circle like a snow globe, and the ellipse settles to the bottom under gravity, assume with frictionless ellipse-circle contact.
Q. For which $a$ and $b$ will the ellipse settle to a single point of contact?
I would especially appreciate an answer that avoids excessive calculations.
You need the curvature radius of the ellipse at $\dfrac{\pi}{2}$ to be smaller than the curvature radius of the circle.
That is:
$\dfrac{a^2}{b}<1$
EDIT: the general expression of the curvature radius of an ellipse of equation $x=a \cos t$, $y=b \sin t$ is $R_c=\dfrac{a^2}{b}(1-e^2\cos^2 t)^{\frac 32}$, with $e=\sqrt{1-(\frac{b}{a})^2}$ being the excentricity of the ellipse.
You have also directly from this that the radius of curvature at $0$ is $\dfrac{b^2}{a}$