 Solution

Put $s^2=x^2+y^2$. For $\rho >0$, elimination of $u$ givesThese are oblate spheroids with equatorial radius $\cosh \rho$ and polar radius $\sinh \rho$, flattened along the $z$ axis. At $\rho =0$ the limiting surface is the disk $z=0$, $s\le 1$.

For $0<|u|<\pi /2$, elimination of $\rho$ instead givesThese are one-sheet hyperboloids of revolution, restricted to their upper half for $u>0$ and lower half for $u<0$. The restriction matters because $\rho \ge 0$. At $u=0$, the surface degenerates to the exterior equatorial plane $z=0$, $s\ge 1$; at $u=\pm \pi /2$ it degenerates to the corresponding half of the $z$ axis.

Finally, fixed $v$ specifies a vertical half-plane through the $z$ axis, with horizontal direction $(\cos v,\sin v)$. The two values $v$ and $v+\pi$ together give a full plane. These surfaces form the mutually orthogonal families of the hyperbolic parametrization of oblate spheroidal coordinates.