In the $\mathbb{H}^3$ upper half space model, is a hemiellipsoid perpendicular to the plane at infinity a minimal surface? 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).
 A: On p. 21 of Polthier's thesis, equation (2.1), you'll find the explicit minimal surface equation in the upper half-space model. I plugged the formula for an ellipsoid into this equation in Mathematica (I'm too lazy to do the computation), and found that the only solutions are for hemispheres. I've included a screenshot from the Mathematica notebook: the appropriately scaled form of the differential equation is a quadratic form which must be identically zero. The only solution which makes the coefficients zero is when the ellipsoid is a sphere. 

A: It is clear that the triaxial hemiellipsoid is not a minimal surface in general.  Indeed, consider the special case the vertical principal axis is the same as one of the axes in the plane at infinity (of course I am comparing the Euclidean lengths of the axes).  In other words, this is an ellipsoid of revolution.  Then at the point of the ellipsoid on the vertical axis, one of the principal directions is a geodesic and therefore the principal curvature vanishes, whereas the other direction is clearly not a geodesic (unless the ellipsoid is a hemisphere). Thus the mean curvature at this point is nonzero.
A: An ellipsoid half is not an isocurve, and therefore can not represent an isocurve in a conformal mapping.  Since a minimal surface is an isocurve (in any geometry), an ellipsoid (including those at infinity), is not a minimal surface.  Only a hemisphere is.
The base projection of this projection is a Möbius plane, where every circle and straight line represents a 'straight line'.  The Möbius plane is the geometry of the plane at infinity of both the H3 and E3.  It is just that in E3 you don't move far enough to see any appreciable difference.
The curvature of the plane at infinity is zero.  But hyperbolic geometry has a negative curvature, which means the euclidean plane (and hence the Möbius plane, which has the geometry of the complete euclidean plane), is a curve.  
The centre of this projection is the point opposite (ie the zenith), which is also at infinity.  One of the interesting features of the Möbius plane, is that it's a conformal mapping of the Euclidean plane, if you suppose any point U is the point at infinity, and a straight line is a circle passing through U.  When U is the zenith point, the conformal projection of the Euclidean plane at the base is the euclidean plane in nature.
