= Solution
The <proper time> normalization of the astronaut's <four-velocity> gives $\dot t^2-\dot x^2=1$. In the <light-cone coordinates> this is
$$
\boxed{\dot p\dot q=1.}
$$
The <four-acceleration> has squared norm $\ddot t^2-\ddot x^2=-a^2$, because $a$ is the constant <proper acceleration>. The same coordinate change therefore gives
$$
\boxed{\ddot p\ddot q=-a^2.}
$$
Equivalently write $\dot q=e^\eta$, $\dot p=e^{-\eta}$. Their second derivatives give $\ddot p\ddot q=-\dot\eta^2$, so the <rapidity form of one-dimensional proper acceleration> makes $\dot\eta=\pm a$, with sign fixing the acceleration direction.
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