Slices of an ellipsoid are ellipses: Can you slice an oblate ellipsoid to get an ellipse that is more oblate?

Question

I'm struggling visualizing this. Is it possible to slice an ellipsoid with minor to major axis = $F$ and get an ellipse with axis ratios less than $F$?

Answer 1

Without loss of generality we may assume that the major semi-axis of your ellipsoid is $1$ and minor semi-axis is $a<1$ (hence $F=a$), so we can write the equation of the ellipsoid as $$ \tag{1} x^2+y^2+{z^2\over a^2}=1. $$ We are free to choose $x$ and $y$ axes such that the intersecting plane is parallel to $y$ axis. Planes parallel to $z$ axis give rise to ellipses with axis ratio equal to $a$, hence we may ignore them. The equation of the intersecting plane can then be written as $$ \tag{2} z=mx+k. $$ The intersection between ellipsoid and plane is an ellipse, whose minor axis is a line on $xz$ plane having the same equation $(2)$ as the plane. The intersections $A$ and $B$ between this line and the ellipsoid can then be found plugging $y=0$ and $z=mx+k$ into $(1)$ and solving the equation for $x$. Their $x$ coordinates are then given by $$ x={-mk\pm\sqrt{m^2k^2-(a^2+m^2)(k^2-a^2)}\over a^2+m^2} $$ and their $z$ coordinates can be found from $(2)$. We can then compute distance $AB$, which is the length of the minor axis of the ellipse: $$ AB={2a\sqrt{1+m^2}\sqrt{a^2+m^2-k^2}\over a^2+m^2}. $$ The center of the ellipse is the midpoint $M$ of $AB$: $$ x_M=-{mk\over a^2+m^2},\quad z_M={a^2k\over a^2+m^2}. $$ The major axis of the ellipse is the line through $M$ parallel to $y$ axis. Its intersections with the ellipse can be found plugging $x$ and $z$ coordinates of $M$ into $(1)$ and solving for $y$: $$ y^2={a^2+m^2-k^2\over a^2+m^2}. $$ From that, we can find the length of major axis $CD$: $$ CD=2{\sqrt{a^2+m^2-k^2}\over \sqrt{a^2+m^2}}. $$ Now we can compute the desired ratio: $$ {AB\over CD}=a{\sqrt{1+m^2}\over \sqrt{a^2+m^2}}. $$ As $a<1$ this ratio is always greater than $a$, as it was to be proved.