= Solution
For $a>0$, the positive constant-$u$ surfaces in <elliptic cylindrical coordinates> satisfy
$$
\frac{x^2}{a^2\cosh^2u}+\frac{y^2}{a^2\sinh^2u}=1.
$$
They are confocal <elliptic cylinders> with focal lines $x=\pm a,y=0$. For a generic fixed $v$, eliminating $u$ gives
$$
\frac{x^2}{a^2\cos^2v}-\frac{y^2}{a^2\sin^2v}=1.
$$
Because $u\geq0$, a fixed $v$ selects the half-branch with the signs of $x$ and $y$ fixed by $\cos v$ and $\sin v$. These are <hyperbolic cylinders>. The degenerate $u=0$ surface is $y=0$, $|x|\leq a$, extended in $z$. At $v=0,\pi$ the cross-sections are the two exterior horizontal rays; at $v=\pi/2,3\pi/2$ they are vertical half-axes.
\Image[/past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2003/paper-1-elliptic-coordinate-net.png]
{title=Confocal elliptic and hyperbolic coordinate curves}
{description=Original cross-section with focal points marked. Ellipses are constant-u curves and the hyperbola half-branches are constant-v curves; extruding in z gives their coordinate surfaces.}
Differentiate the position <vector> to get the coordinate <tangent vectors>
$$
r_u=(a\sinh u\cos v,a\cosh u\sin v,0),\quad r_v=(-a\cosh u\sin v,a\sinh u\cos v,0),\quad r_z=(0,0,1).
$$
Their pairwise <dot products> vanish. Their squared lengths are $a^2(\sinh^2u+\sin^2v)$, $a^2(\sinh^2u+\sin^2v)$ and one. This proves <orthogonality> and gives the <scale factors of orthogonal coordinates>:
$$
\boxed{h_u=h_v=h=a\sqrt{\sinh^2u+\sin^2v},\qquad h_z=1.}
$$
The <Riemannian metric> is $ds^2=h^2(du^2+dv^2)+dz^2$. Away from the focal lines, divide the coordinate <tangent vectors> by their lengths to obtain the local orthonormal <unit vectors> $e_u,e_v,e_z$. For physical <vector field> components $A_u,A_v,A_z$, the formulas for <vector calculus in elliptic cylindrical coordinates> are
$$
\boxed{\nabla\cdot A=\frac{\partial_u(hA_u)+\partial_v(hA_v)+\partial_z(h^2A_z)}{h^2},}
$$
$$
\boxed{\nabla\times A=\frac{\partial_vA_z-\partial_z(hA_v)}h e_u+\frac{\partial_z(hA_u)-\partial_uA_z}h e_v+\frac{\partial_u(hA_v)-\partial_v(hA_u)}{h^2}e_z.}
$$
The <gradient> of a <scalar> <function> has components $(\phi_u/h,\phi_v/h,\phi_z)$. Apply the <divergence> formula to this <gradient>; <Laplace's equation> becomes
$$
\boxed{\frac{\phi_{uu}+\phi_{vv}}{a^2(\sinh^2u+\sin^2v)}+\phi_{zz}=0.}
$$
The formulas apply where $h>0$; the focal lines require a limit or another coordinate chart.
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