Answer
Using the results from Exercise 13:
${r_{per}} = a\left( {1 - e} \right)$ and ${r_{ap}} = a\left( {1 + e} \right)$,
we prove that
$e = \frac{{{r_{ap}} - {r_{per}}}}{{{r_{ap}} + {r_{per}}}}$
$p = \frac{{2{r_{ap}}{r_{per}}}}{{{r_{ap}} + {r_{per}}}}$
Work Step by Step
From Exercise 13, we obtain ${r_{per}} = a\left( {1 - e} \right)$ and ${r_{ap}} = a\left( {1 + e} \right)$.
It follows that $a = \frac{{{r_{per}}}}{{1 - e}}$. Substituting it in ${r_{ap}}$ gives
${r_{ap}} = \frac{{{r_{per}}\left( {1 + e} \right)}}{{1 - e}}$
${r_{ap}} - e{r_{ap}} = {r_{per}} + e{r_{per}}$
${r_{ap}} - {r_{per}} = e\left( {{r_{ap}} + {r_{per}}} \right)$
Hence, $e = \frac{{{r_{ap}} - {r_{per}}}}{{{r_{ap}} + {r_{per}}}}$.
From Exercise 13, we obtain $p = a\left( {1 + e} \right)\left( {1 - e} \right)$.
Substituting $a = \frac{{{r_{per}}}}{{1 - e}}$ and $e = \frac{{{r_{ap}} - {r_{per}}}}{{{r_{ap}} + {r_{per}}}}$ in $p$ gives
$p = \frac{{{r_{per}}}}{{1 - \frac{{{r_{ap}} - {r_{per}}}}{{{r_{ap}} + {r_{per}}}}}}\left( {1 + \frac{{{r_{ap}} - {r_{per}}}}{{{r_{ap}} + {r_{per}}}}} \right)\left( {1 - \frac{{{r_{ap}} - {r_{per}}}}{{{r_{ap}} + {r_{per}}}}} \right)$
$p = {r_{per}}\left( {1 + \frac{{{r_{ap}} - {r_{per}}}}{{{r_{ap}} + {r_{per}}}}} \right)$
$p = {r_{per}}\left( {\frac{{2{r_{ap}}}}{{{r_{ap}} + {r_{per}}}}} \right)$
Hence, $p = \frac{{2{r_{ap}}{r_{per}}}}{{{r_{ap}} + {r_{per}}}}$.
