The Kepler orbit equation is a conic: $u=c(1+e\cos (\phi -{\theta }_0))$ with $E=(\mu {\kappa }^2\mathrm{/}2J^2)(e^2-1)$

Note $u^{\mathrm{\prime }\mathrm{\prime }}+u=c$ is equivalent to $(u-c{)}^{\mathrm{\prime }\mathrm{\prime }}+(u-c)=0$ on $W$ by linearity of the second derivative. Since $W$ is an interval and Binet's equation holds on it (it contains at least two points because $I^{\circ }$ has at least two points), [L2] gives reals $A,B$ with $u(\phi )-c=A\cos (\phi -{\phi }_0)+B\sin (\phi -{\phi }_0)(\phi \in W)$ for a fixed ${\phi }_0\in W$.

Put $R:=\sqrt{A^2+B^2}\ge 0$. If $R>0$, then $(A\mathrm{/}R,B\mathrm{/}R)$ is a unit vector, so by [L3] there is $\psi \in [0,2\pi )$ with $\cos \psi =A\mathrm{/}R$, $\sin \psi =B\mathrm{/}R$; if $R=0$ take $\psi :=0$. Then for every $\phi$, [L3] gives $A\cos (\phi -{\phi }_0)+B\sin (\phi -{\phi }_0)=R\cos (\phi -{\phi }_0-\psi )\mathrm{.}$ Setting ${\theta }_0:={\phi }_0+\psi$ and $e:=R\mathrm{/}c=R{J}^2\mathrm{/}(\mu \kappa )\ge 0$, step 1.1 becomes $u(\phi )=c+ec\cos (\phi -{\theta }_0)=c(1+e\cos (\phi -{\theta }_0))$ on $W$, which is the first assertion; the polar form $\rho =p\mathrm{/}(1+e\cos (\theta -{\theta }_0))$ follows by inverting $u=1\mathrm{/}\rho$ on $I^{\circ }$.

Energy: by [L1] and step 2.1, with $\psi (\phi ):=\phi -{\theta }_0$, $u=c(1+e\cos \psi )$ and $u^{\mathrm{\prime }}=-ce\sin \psi$ (differentiate with [L4]); hence $(u^{\mathrm{\prime }}{)}^2+u^2=c^2(e^2{\sin }^2\psi +1+2e\cos \psi +e^2{\cos }^2\psi )=c^2(1+2e\cos \psi +e^2),$ and $\kappa u=\kappa c(1+e\cos \psi )$. Substituting into [L1]: $E=\frac{J^2{c}^2}{2\mu }(1+2e\cos \psi +e^2)-\kappa c(1+e\cos \psi )\mathrm{.}$ With $c=\mu \kappa \mathrm{/}{J}^2$ one has $\frac{J^2{c}^2}{2\mu }=\frac{\mu {\kappa }^2}{2J^2}$ and $\kappa c=\frac{\mu {\kappa }^2}{J^2}$, so $E=\frac{\mu {\kappa }^2}{2J^2}\left[\right(1+2e\cos \psi +e^2)-2(1+e\cos \psi \left)\right]=\frac{\mu {\kappa }^2}{2J^2}(e^2-1)\mathrm{.}$

Classification: since $\mu {\kappa }^2\mathrm{/}(2J^2)>0$, the sign of $E$ equals the sign of $e^2-1$; with $e\ge 0$ this is negative exactly for $e<1$, zero exactly for $e=1$ and positive exactly for $e>1$.