= Solution
Put $s^2=x^2+y^2$. For $\rho>0$, elimination of $u$ gives
$$
\boxed{\frac{s^2}{\cosh^2\rho}+\frac{z^2}{\sinh^2\rho}=1.}
$$
These 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 $z=0$, $s\leq1$.
For $0<|u|<\pi/2$, elimination of $\rho$ instead gives
$$
\boxed{\frac{s^2}{\cos^2u}-\frac{z^2}{\sin^2u}=1.}
$$
These are <one-sheet hyperboloids> of revolution, restricted to their upper half for $u>0$ and lower half for $u<0$. The restriction matters because $\rho\geq0$. At $u=0$, the surface degenerates to the exterior equatorial plane $z=0$, $s\geq1$; 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>.
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