Answer
Using the fact that ${r_{per}} + {r_{ap}} = 2a$, we obtain
${r_{per}} = \frac{p}{{1 + e}} = a\left( {1 - e} \right)$
${r_{ap}} = \frac{p}{{1 - e}} = a\left( {1 + e} \right)$
Work Step by Step
At $\theta=0$, we obtain the distance from the sun at the perihelion:
${r_{per}} = \frac{p}{{1 + e}}$
Similarly, at $\theta = \pi $, we obtain the distance from the sun at the aphelion:
${r_{ap}} = \frac{p}{{1 - e}}$
Using the fact that ${r_{per}} + {r_{ap}} = 2a$, so
$\frac{p}{{1 + e}} + \frac{p}{{1 - e}} = 2a$
$p\left( {\frac{1}{{1 + e}} + \frac{1}{{1 - e}}} \right) = 2a$
$p\left( {\frac{2}{{\left( {1 + e} \right)\left( {1 - e} \right)}}} \right) = 2a$
$p = a\left( {1 + e} \right)\left( {1 - e} \right)$
Substituting $p$ in ${r_{per}}$ and ${r_{ap}}$ gives
${r_{per}} = \frac{p}{{1 + e}} = a\left( {1 - e} \right)$
${r_{ap}} = \frac{p}{{1 - e}} = a\left( {1 + e} \right)$
