Conjugate diameters of ellipsoid

Question

I am familiar with ellipses,but am seeking to find the condition for 2 diameters of an ellipsoid to be conjugate.Any help or references would be appreciated.

Answer 1

Here's the definition of conjugate diameters for an ellipsoid (from "Solid Geometry" by W.H. Macauley):

Three diameters of an ellipsoid, such that the plane through any two of them bisects all chords parallel to the third, are called a set of conjugate diameters.

In practice, given any diameter of the ellipsoid, one can consider the conjugate plane passing through the center of the ellipsoid and parallel to the tangent planes at the endpoints of the given diameter. The intersection of the conjugate plane with the ellipsoid is an ellipse, and any two conjugate diameters of this ellipse, together with the given diameter, form a set of conjugate diameters of the ellipsoid.

EDIT.

It is easy to prove that any three equal diameters of an ellipsoid (not necessarily conjugated!) lie on a right circular cone. Let $O$ be the center and $OA$, $OB$, $OC$ three radii of the ellipsoid: points $A$, $B$, $C$ lie on a sphere centred at $O$ and as a consequence the center of their circumcircle is the projection of $O$ on plane $ABC$. Hence those radii lie on a right cone of vertex $O$ and base circle $ABC$.

It is not clear what is meant by "to determine the angle between any 2 of them", because infinite solutions are possible. If $a$, $b$, $c$ are the semi-axes of the ellipsoid, then we know that the endpoints $A$, $B$, $C$ of three equal conjugate diameters lie on a sphere with radius $\sqrt{(a^2+b^2+c^2)/3}$. The intersection of that sphere with the ellipsoid is curve (not planar, in general) with two branches (symmetric about $O$). Taken $A$ on that curve, we can find the plane conjugate to $OA$, which cuts the same branch of the curve at $B$ and $C$: it is not difficult to see that $OA$, $OB$ and $OC$ are a set of conjugate semi-diameters for the ellipsoid, but the angles among them vary, in general, with the position of $A$.

EDIT 2.

In reading again your comment, where you mention a spheroid, I think you are considering an ellipsoid with rotational symmetry, with semi-axes $a=b$ and $c$. In this case the three ellipses corresponding to any couple of conjugate diameters are equal by symmetry, and the angles $\theta$ between conjugate diameters are all equal among them.

In this case all conjugate semi-diameters have lengths $\sqrt{(2a^2+c^2)/3}$. Consider an ellipse having two of the conjugate diameters of the spheroid as conjugate diameters. If $\alpha$ and $\beta$ are the semi-axes of that ellipse, then: $$ \alpha^2+\beta^2=2{2a^2+c^2\over3} \quad\text{and}\quad \alpha\beta={2a^2+c^2\over3}\sin\theta. $$ But by symmetry one of the semi-axes of the ellipse must be the same as the "equatorial" semi-axis of the spheroid, i.e. $\alpha=a$. It follows that $\beta=\sqrt{(2c^2+a^2)/3}$ and finally: $$ \sin\theta={3\alpha\beta\over 2a^2+c^2}= \sqrt3\ {a\sqrt{2c^2+a^2}\over 2a^2+c^2}. $$