 Solution

The tangent planecuts the hyperboloid inso its intersection is the pair of straight linesTheir accelerations vanish, hence are normal to the surface, so both lines are geodesics. Their tangent vectors at their intersection $(1,0,0)$ are $(0,1,1)$ and $(0,-1,1)$, whose inner product is zero. They therefore intersect at a right angle, giving the second case of plane-section geodesics of the unit one-sheet hyperboloid.

There are also geodesics entirely contained in $z>0$. In the coordinatesthe metric isFor a unit-speed geodesic, the Clairaut first integral for a surface of revolution is $c=(1+z^2)\dot{\varphi }$, and the constant-speed equation becomesChoose $c>1$ and initial height $z_0=\sqrt{c^2-1}$. The geodesic initially tangent to the parallel has $\dot{z}=0$, and the displayed identity prevents it from entering $0<z<z_0$. Since the parallel at $z_0>0$ is not itself a geodesic, the curve turns there and otherwise has $z>z_0$. Hence it remains entirely in $z>0$; this is a geodesic trapped in one half of the unit one-sheet hyperboloid.

Finally, the waistis a geodesic and is preserved setwise by every isometry of the hyperboloid. Indeed, its Gaussian curvature isThe value $K=-1$ occurs exactly at $z=0$. Since Gaussian curvature is intrinsic and therefore preserved by isometries, every isometry preserves the waist. Thus the answer to the final question is also yes, with the isometry-invariant waist geodesic of the unit one-sheet hyperboloid as an example.