= Solution
The plane
$$
\boxed{y=0}
$$
meets the <one-sheet hyperboloid> in the two disjoint curves
$$
(x,y,z)=(\pm\sqrt{1+z^2},0,z).
$$
They are the two opposite meridians of the <surface of revolution>. Parametrize the generating profile by <arc length>. Its <acceleration> lies in the meridian plane and is orthogonal to the profile tangent; it is also orthogonal to the azimuthal tangent. It is therefore normal to the surface, so both meridians are geodesics. This is the disjoint case in <plane-section geodesics of the unit one-sheet hyperboloid>.
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