 Pericentre-to-crossover flight time

A parabolic Kepler orbit obeys $t=\sqrt{2q^3/(GM)}(D+D^3/3)$ with $D=\tan (f/2)$. At the high-eccentricity angular-speed crossover, $q\ll r_x$, so $n_p{t}_x\simeq (\sqrt{2}/3)(r_x/a_p{)}^{3/2}=2^{7/8}(q/a_p{)}^{3/8}/3$. The small time fraction explains why an eccentric orbit turns through almost $\pi$ of true anomaly while a distant perturber barely moves.