Use geometric optics and a property of a parabola. Parallel rays, directed along the axis of the mirror, intersect at the focus of the parabola. The same is true in 2 dimensions with a cylindrically symmetric paraboloid (that's because you can always look at a vertical 2D cross section that includes the ray and the axis of the paraboloid).
This means that any ray along (0,0,-1) direction will go through the focus, which lies on the symmetry axis of the paraboloid.
The focus of the parabola $z=r^2$, where I simply introduced $r$ as distance from the axis, is at $z=\frac{1}{4}$.
Verify this by finding $r$ at which the derivative of $z=r^2$ is $z'=2r=1\rightarrow r=\frac12$. At that point, the mirror is exactly at $45^\circ$ and reflects a vertical ray into the horizontal direction. This ray therefore goes through $z=r^2=\left(\frac{1}{2}\right)^2=\frac14$.
This is not necessary if you know where the focus of a parabola is (recall the canonical form $x^2=4py$).
Use geometric optics and a property of a parabola. Parallel rays, directed along the axis of the mirror, intersect at the focus of the parabola. The same is true in 2 dimensions with a cylindrically symmetric paraboloid (that's because you can always look at a vertical 2D cross section that includes the ray and the axis of the paraboloid).
This means that any ray along (0,0,-1) direction will go through the focus, which lies on the symmetry axis of the paraboloid.
The focus of the parabola $z=r^2$, where I simply introduced $r$ as distance from the axis, is at $z=\frac{1}{4}$.
Verify this by finding $r$ at which the derivative of $z=r^2$ is $z'=2r=1\rightarrow r=\frac12$. At that point, the mirror is exactly at $45^\circ$ and reflects a vertical ray into the horizontal direction. This ray therefore goes through $z=r^2=\left(\frac{1}{2}\right)^2=\frac14$.
This is not necessary if you know where the focus of a parabola is (recall the canonical form $x^2=4py$).