= Solution
Use the specified hyperbolic trajectory. The emitted right-moving <photon> follows $x=L+t-t_0$, so its retarded <null coordinate> is $p=t_0-L$. Interception at astronaut <proper time> $\tau$ requires
$$
t_0-L=-\frac1a e^{-a\tau}.
$$
The astronaut has velocity $v=\tanh(a\tau)$ relative to the star and $\gamma=\cosh(a\tau)$. The collinear <relativistic Doppler effect> gives
$$
\boxed{\nu_{\rm obs}=\nu_0\gamma(1-v)=\nu_0e^{-a\tau}=a\nu_0(L-t_0).}
$$
This is the <Doppler shift for a uniformly accelerated receiver>. A finite reception time exists precisely when $t_0<L$, so \b[the limiting emission time is $L$], approached from below. At $t_0=L$ the photon lies on the <Rindler horizon> and never reaches the astronaut. If reception is restricted to $\tau\ge0$, the full emission range is $L-1/a\le t_0<L$; for emission after $t=0$, it is $0\le t_0<L$.
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