 Velocity kick on a circular Kepler orbit

For an instantaneous in-plane change of velocity of magnitude $\gamma v_K$ on a circular orbit, with angle $\theta$ measured from the forward tangential direction, the new specific orbital energy gives $a^{\prime }/a=(1-2\gamma \cos \theta -{\gamma }^2{)}^{-1}$. The specific angular momentum gives $e^{\prime 2}=1-(1-2\gamma \cos \theta -{\gamma }^2)(1+\gamma \cos \theta {)}^2$. The signed semi-major axis becomes negative for an unbound hyperbolic Kepler orbit.