 Solution

For the stated outward-accelerating hyperbola, take positive signed proper acceleration and an initially stationary spaceship. Then $\eta =a\tau$, so $\dot{p}=e^{-a\tau }$ and $\dot{q}=e^{a\tau }$. The initial values are $p(0)=-1/a$, $q(0)=1/a$. Integration givesTaking their half-sum and half-difference yields the hyperbolic motionThe printed position data alone do not specify the initial velocity. If the initial rapidity is an arbitrary ${\eta }_0$, the equally valid positive-acceleration solution isThus the requested special expression implicitly selects ${\eta }_0=0$.