Yes, you can take $a$ and $c$ as actual distances. The perihelion is $d_p=a-c$ and the aphelion is $d_a=a+c$, so $$d_a=2c+d_p=2ea+d_p\\\d_a=2a-d_p\\\ed_a=2ae-ed_p\\\\(1-e)d_a=(1+e)d_p\\\d_a=\frac {1+e}{1-e}d_p$$
Yes, you can take $a$ and $c$ as actual distances. The perihelion is $d_p=a-c$ and the aphelion is $d_a=a+c$, so $$d_a=2c+d_p=2ea+d_p\\\d_a=2a-d_p\\\ed_a=2ae-ed_p\\\\(1-e)d_a=(1+e)d_p\\\d_a=\frac {1+e}{1-e}d_p$$