Calculating surface of ellipsoid visible from any point outside of it

Getting the boundary of the visible region is much easier than calculating its area. The boundary curve is an ellipse. It is the intersection of the ellipsoid and the so-called polar plane of the view-point.

If the ellipsoid has equation $$ \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1 $$ Then the polar plane of the point $(u,v,w)$ has equation $$ \frac{ux}{a^2} + \frac{vy}{b^2} + \frac{wz}{c^2} = 1 $$ And, actually, I have now found that everything you need is probably in this paper. The details of the ellipse “horizon” are worked out, and the enclosed area is computed by numerical methods.

Assuming you are using an actual spheroid (which WGS84 is as far as I understand) and not some other kind of geoid, your spheroidal geoid can be mapped to a sphere via an affine transformation that "stretches" the Earth along the polar axis.

An affine transformation maps lines that are tangent to the spheroid to lines that are tangent to the sphere, and maps concurrent lines to concurrent lines, so whatever the lines of sight are to the "horizon" from the position of your satellite, they map to the lines tangent to the sphere from whatever point in space the affine transformation takes your satellite to.

Since the horizon on the sphere is a circle in a plane, the horizon on the spheroid (after reversing the affine transformation) is also in a plane, though it is not generally a circle since that shape is not preserved.

The horizon will be an ellipse, which seems less ugly than the GIS.se answer makes it out to be. They may have been thinking of the geodetic curves on the spheroid, which do not map to geodetics on the sphere under the affine transformation. In addition, the lines of latitude are not preserved by the affine transformation either, so the determination of latitudes and longitudes along the horizon may require an extra calculation.

The real problem is that relative surface areas are not preserved by the affine transformation between a sphere and a spheroid, so to get the total visible area you would have to integrate an ellipsoidal surface within an arbitrary elliptical boundary. There is not even a closed formula for the length of an arc of an ellipse using anything more commonplace than elliptical functions, so I'd be surprised if there is a simple closed formula for the area you want.

Perhaps someone reading this answer will recognize the appropriate integral to set up in some kind of "closed form" (though perhaps using functions that most people know little about).