Given a Jordan curve on the $\mathbb{H}^3$ boundary at infinity, there is a minimal surface (topological disk) in $\mathbb{H}^3$ with the curve as its asymptotic boundary (page.mi.fu-berlin.de/polthier/articles/diss/diss.pdf, Theorem 21 on p37).

Consider boundary curves in the upper half space model. If the boundary curve is a circle, the minimal surface is a geodesic plane (a hemisphere orthogonal to the boundary). A hemiellipsoid orthogonal to the boundary (a triaxial ellipsoid with two axes in the plane at infinity, and the third axis vertical) has a boundary curve that is an ellipse, and it seems likely that a hemiellipsoid is the minimal surface for this boundary curve. Why would it be something more complicated?

For the point on the hemiellipsoid through its vertical axis, it is easy to see that one (hyperbolic) principle curvature is positive and the other is negative, by considering the hemisphere with the same vertical axis through that point. But it would be nice to have an argument that the hyperbolic principle curvatures are equally opposite (*update:* based on comments/answers, we should be able to pick the ellipsoid parameters to make this true at this specific point), and that this is true for all points on the hemiellipsoid.

The motivation for this question comes from $\mathbb{H}^3$ fibrations, specifically about lifts of geodesics from the base $\mathbb{H}^2$ surface. For some fibrations, the lift is a hemiellipsoid in the upper half space model (see plus.google.com/+RoiceNelson/posts/1w3aoQgj61g and comments therein).