Answer
We evaluate the magnitude $J$ at the perihelion and aphelion. Then, using the result from Exercise 13 we prove that
${v_{per}}\left( {1 - e} \right) = {v_{ap}}\left( {1 + e} \right)$
Work Step by Step
Since ${\bf{J}} = {\bf{r}} \times {\bf{r}}'$ is constant and ${\bf{r}}$ is perpendicular to ${\bf{r}}'$ at the perihelion and aphelion; for convenience we evaluate the magnitude $J$ at the perihelion and aphelion:
$||{\bf{J}}|| = ||{{\bf{r}}_{per}}||||{{\bf{r}}_{per}}'||\sin \frac{\pi }{2} = {r_{per}}{v_{per}}$
$||{\bf{J}}|| = ||{{\bf{r}}_{ap}}||||{{\bf{r}}_{ap}}'||\sin \frac{\pi }{2} = {r_{ap}}{v_{ap}}$
So,
(1) ${\ \ \ \ }$ ${r_{per}}{v_{per}} = {r_{ap}}{v_{ap}}$
From Exercise 13, we obtain ${r_{per}} = \frac{p}{{1 + e}}$ and ${r_{ap}} = \frac{p}{{1 - e}}$. Substituting these in equation (1) gives
$\frac{p}{{1 + e}}{v_{per}} = \frac{p}{{1 - e}}{v_{ap}}$
Hence, ${v_{per}}\left( {1 - e} \right) = {v_{ap}}\left( {1 + e} \right)$.
