= Solution
\b[Coordinate surfaces and sketch.] Adding and subtracting the defining relations gives $s=u^2+v^2$ and $x=u^2-v^2$. Since $y^2=s^2-x^2$, we have $y^2=4u^2v^2$. For $u^2=c>0$, eliminate $v^2$ to obtain
$$
x=c-\frac{y^2}{4c}.
$$
This is a <parabolic cylinder> parallel to the $z$ axis, opening towards negative $x$, with cross-sectional vertex $(c,0)$ and focus at the origin. Similarly, $v^2=d>0$ gives
$$
x=\frac{y^2}{4d}-d,
$$
a <parabolic cylinder> opening towards positive $x$, with vertex $(-d,0)$ and the same focus. The degenerate $u^2=0$ surface is the half-plane $y=0,x\leq0$; $v^2=0$ gives $y=0,x\geq0$. Constant $w$ gives the horizontal <planes> $z=w$.
Thus the requested cross-sections consist of $x=1-y^2/4$, $x=2-y^2/8$, $x=y^2/4-1$, $x=y^2/8-2$ and the two horizontal half-axes. The <parabolas> of the two families meet orthogonally.
\Image[/past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2002/paper-1-parabolic-coordinate-net.png]
{title=The requested parabolic coordinate curves in the xy plane}
{description=Blue curves have u squared equal to one or two; orange curves have v squared equal to one or two. The thick horizontal rays are the degenerate zero levels. All parabolas share the origin as focus.}
\b[Signed coordinates.] Take $u\geq0$ and choose the sign of $v$ to agree with $y$ away from $y=0$. Then the <Cartesian coordinates> are
$$
\boxed{x=u^2-v^2,\qquad y=2uv,\qquad z=w.}
$$
Every coordinate triple gives exactly one point. For $u>0$, $v$ and $y$ have the same sign. A fixed signed $v\ne0$ selects the upper or lower half of its <parabolic cylinder>; the entire cylinder corresponds to fixed $v^2$. On the cut $u=0$, the two values $v=\pm\sqrt{-x}$ represent the same point of the negative $x$ half-axis. This boundary identification is unavoidable: there $y=0$ even when $v\ne0$, so the sign prescription is understood by approaching from either side of the cut. The interior $u>0$ gives a unique coordinate chart off that half-axis.
\b[Orthogonality and metric.] Differentiate the <position vector>:
$$
r_u=(2u,2v,0),\qquad r_v=(-2v,2u,0),\qquad r_w=(0,0,1).
$$
The pairwise <dot products> vanish. Their lengths give the <scale factors of orthogonal coordinates> and hence the <Riemannian metric>:
$$
\boxed{h_u=h_v=2\sqrt{u^2+v^2},\quad h_w=1,\qquad dl^2=4(u^2+v^2)(du^2+dv^2)+dw^2.}
$$
These are <parabolic cylindrical coordinates>. The differential calculation is regular for $u^2+v^2>0$; the $z$ axis is the coordinate singularity where the two planar <tangent vectors> vanish.
\b[Laplacian.] Put $h=2\sqrt{u^2+v^2}$. The physical components of the <gradient> of a $w$-independent <scalar field> are $\phi_u/h,\phi_v/h,0$. Applying the <divergence> formula cancels the two equal <scale factors of orthogonal coordinates> inside the derivatives:
$$
\nabla^2\phi=\frac1{h^2}\left[\partial_u\left(h\frac{\phi_u}h\right)+\partial_v\left(h\frac{\phi_v}h\right)\right].
$$
Therefore, away from the coordinate singularity,
$$
\boxed{\nabla^2\phi=\frac{\phi_{uu}+\phi_{vv}}{4(u^2+v^2)}.}
$$
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