Past exam of the mathematics course of the University of Cambridge / 2019 / ia / Paper 2 / 8C / c / Solution

The initial point has $H=1/4$. On the branch leaving $(1/2,0)$ toward negative $y$,Since $\dot{y}=-x$, the required scalar equation isIts solution isAt $(x,y)=(1/4,-1/2)$, this requires $\tanh (t/\sqrt{2})=1/\sqrt{2}$. Using the inverse hyperbolic tangent,