Geodesic on hyperboloid and Poincare's Disk Model

Question

I have two questions that

1. Why geodesic on hyperboloid corespond the arc in the Poincare's Disk Model?

The hyperboloid : $x^2 + y^2 - z^2 = -1, \hspace{.15cm} z>0$

When any plane through origin cut the hyperboloid, the intersecting curve is a geodesic, I know.

My second question is that

2. Why the geodesic is a hyperbola?

I want some basic proof, thanks.

Proofs can be found in textbooks, e.g. Ratcliffe. – Lee Mosher Aug 19 '15 at 14:24 — Lee Mosher, Aug 19 '15 at 14:24

score 1 · Answer 1 · answered Aug 09 '15 at 16:19

The correspondence between the Poincare disc model and the hyperbolic model is via the projection from the source at $(0,0,-1)$ point. Suppose that there is a light source at the point $(0,0,-1)$. Then each ray connects an unique point of the unit disc (centered at origin, in the XY plane) to a unique point in the hyperboloid. Now try to picture the projection of a geodesics in the hyperboloid model. You will see that they are the arcs in disc which intersects the boundary perpendicularly (not all arcs).

For your second question, try to think why parabola, hyperbola and ellipse are called conic sections. See the figures in this Wikipedia page.

Thank you. but I want the proof... – c-301 Aug 10 '15 at 04:03 — c-301, Aug 10 '15 at 04:03

Geodesic on hyperboloid and Poincare's Disk Model

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