= Solution
For <radial null geodesics in a de Sitter static patch>, the conserved time quantity and the stated normalization at the origin give $\alpha\dot t=1$. The null condition then gives
$$
0=-c^2\alpha\dot t^2+\alpha^{-1}\dot r^2,\qquad \dot r^2=c^2\alpha^2\dot t^2=c^2.
$$
The emitted outgoing ray selects the positive branch, so
$$
\boxed{\dot r=c,\qquad\dot t=\frac1{1-r^2/r_S^2}.}
$$
The ingoing branch would have $\dot r=-c$ instead. Set the affine parameter to zero at emission. Then $r=c\lambda$, and integrating $dt/dr=1/[c(1-r^2/r_S^2)]$ gives
$$
t=\frac{r_S}{c}\operatorname{artanh}\frac r{r_S},\qquad \boxed{r(t)=r_S\tanh\frac{ct}{r_S}.}
$$
The coordinate time diverges as $r\to r_S$, although the outgoing affine parameter reaches $r_S/c$ there. This is a coordinate-horizon effect: a local stationary observer inside the patch measures radial distance $dr/\sqrt\alpha$ and proper time $\sqrt\alpha\,dt$, whose ratio along the ray is still $c$.
\Image[/past-exam-of-the-general-relativity-and-cosmology-course-of-the-university-of-oxford/2012-de-sitter-null-rays-and-clock-redshift.png]
{title=Radial light rays in static coordinates and the redshift of a stationary emitter}
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