= Solution
For $u>0$, eliminating $v$ gives
$$
\boxed{\frac{x^2}{a^2\cosh^2u}+\frac{y^2}{a^2\sinh^2u}=1.}
$$
Thus constant $u$ surfaces are confocal <elliptic cylinders> parallel to the $z$ axis. The limiting $u=0$ surface is the vertical strip $y=0$, $|x|\leq a$.
For generic $v$, eliminating $u$ gives
$$
\boxed{\frac{x^2}{a^2\cos^2v}-\frac{y^2}{a^2\sin^2v}=1.}
$$
These are <hyperbolic cylinders> with the same focal lines. A specified angular value selects the half-branch with $\operatorname{sgn}x=\operatorname{sgn}(\cos v)$ and $\operatorname{sgn}y=\operatorname{sgn}(\sin v)$, because $u\geq0$. At $v=0,\pi$ the surfaces are the vertical half-planes $y=0$, $x\geq a$ or $x\leq-a$; at $v=\pi/2,3\pi/2$ they are $x=0$, $y\geq0$ or $y\leq0$.
Finally, constant $z$ gives horizontal planes. The three surface families express the <orthogonality> of <elliptic cylindrical coordinates>.
Back to article page
