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For $a,b>0$, this quadric surface is independent of its third coordinate and is formed by translating a hyperbola along that direction. Its two connected components correspond to $u\ge a$ and $u\le -a$. In elliptic cylindrical coordinates, fixing a generic angular coordinate selects a half-branch of such a cylinder because the signs of both planar coordinates are fixed by that angular value.

For $a>0$, the positive constant-$u$ surfaces in elliptic cylindrical coordinates satisfyThey are confocal elliptic cylinders with focal lines $x=\pm a,y=0$. For a generic fixed $v$, eliminating $u$ givesBecause $u\ge 0$, 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 $y=0$, $|x|\le a$, 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.

Differentiate the position vector to get the coordinate tangent vectorsTheir pairwise dot products vanish. Their squared lengths are $a^2({\sinh }^{2}u+{\sin }^{2}v)$, $a^2({\sinh }^{2}u+{\sin }^{2}v)$ and one. This proves orthogonality and gives the scale factors of orthogonal coordinates:The Riemannian metric is $d{s}^2=h^2(d{u}^2+d{v}^2)+d{z}^2$. 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 areThe gradient of a scalar function has components $({\varphi }_u/h,{\varphi }_v/h,{\varphi }_z)$. Apply the divergence formula to this gradient; Laplace's equation becomesThe formulas apply where $h>0$; the focal lines require a limit or another coordinate chart.

Differentiate the position vector, keeping the other coordinates fixed in each derivative:The first two have zero dot product, since their two planar contributions cancel, and both are orthogonal to ${\mathbf{h}}_z$. Hence these are orthogonal coordinates away from coordinate degeneracies. Their scale factors, which are the lengths of these coordinate basis vectors, areFor example $|{\mathbf{h}}_u{|}^2=a^2[{\sinh }^{2}u{\cos }^{2}v+{\cosh }^{2}u{\sin }^{2}v]=a^2({\sinh }^{2}u+{\sin }^{2}v)$. The unit basis vectors are ${\mathbf{h}}_i/h_i$ when $h_i\ne 0$; the given coordinate basis vectors themselves are not normalized. The Jacobian determinant is $a^2({\sinh }^{2}u+{\sin }^{2}v)>0$ except on the focal lines. These are elliptic cylindrical coordinates.

For $u>0$, eliminating $v$ givesThus constant $u$ surfaces are confocal elliptic cylinders parallel to the $z$ axis. The limiting $u=0$ surface is the vertical strip $y=0$, $|x|\le a$.

For generic $v$, eliminating $u$ givesThese are hyperbolic cylinders with the same focal lines. A specified angular value selects the half-branch with $\mathrm{sgn}x=\mathrm{sgn}(\cos v)$ and $\mathrm{sgn}y=\mathrm{sgn}(\sin v)$, because $u\ge 0$. At $v=0,\pi$ the surfaces are the vertical half-planes $y=0$, $x\ge a$ or $x\le -a$; at $v=\pi /2,3\pi /2$ they are $x=0$, $y\ge 0$ or $y\le 0$.

Finally, constant $z$ gives horizontal planes. The three surface families express the orthogonality of elliptic cylindrical coordinates.

In elliptic cylindrical coordinates, write a vector field as $A=A_u{e}_u+A_v{e}_v+A_z{e}_z$ using the local orthonormal unit vectors. The scale factors of orthogonal coordinates are $(h,h,1)$, so the divergence and curl areApplying the divergence to the gradient of a scalar function gives $\Delta \varphi =({\varphi }_{uu}+{\varphi }_{vv})/h^2+{\varphi }_{zz}$. These formulas hold where $h>0$. They use physical components along unit vectors, rather than components along the unnormalized coordinate tangent vectors.