Solution
= Solution
At the <periapsis> $r_1$ and <apoapsis> $r_2$, the given <Kepler orbit> satisfies
$$
r_1=\frac{h^2}{k(1+e)},
\qquad
r_2=\frac{h^2}{k(1-e)}.
$$
Hence
$$
e=\frac{r_2-r_1}{r_1+r_2},
\qquad
h^2=\frac{2kr_1r_2}{r_1+r_2}.
$$
Therefore
$$
\boxed{h=\sqrt{\frac{2kr_1r_2}{r_1+r_2}}},
\qquad \boxed{C=2}.
$$