Your two examples are surfaces, not curves. In fact (roughly speaking) any implicit equation of the form $f(x,y,z)=0$ will yield a surface.
One arch-like surface that will work is a parabolic one related to the previous answer:
$ z = 30 \left(1-\frac{y^2}{7225}\right)$
This is an arch-shaped surface that runs parallel to the x-axis. Similarly
$ z = 30 \left(1-\frac{x^2}{7225}\right)$ is an arch-shaped surface that runs parallel to the x-axis. Finally, the arch surface
$z = 30 \left(1-\frac{(x+y)^2}{28900}\right)$
is somehow "nicer" because it's symmetric in $x$ and $y$.
Taking the intersection of any of these three with the plane $x=y$ will give a curve that meets your specifications.