Past exam of the mathematics course of the University of Cambridge / 2020 / ia / Paper 2 / 4C / ii / Solution

Now $E=u^2/2-k/R<0$, so the orbit is bound. Since $L=ub$, the initial energy isAt the farthest central-force radial turning point $q$, the radial speed vanishes and the transverse speed is $L/q$, givingEquating these expressions and multiplying by $2q^2$ yieldsThus $A=L^2/b^2-2k/R$, $B=2k$, and $C=-L^2$.