Answer
$7.746\times10^{9}$km
$1.291\times10^{10}$km
Work Step by Step
Step 1. Identify the given quantities: eccentricity $e=0.25$ and length of minor axis $2b=10$billion km, so that $b=5$ billion km
Step 2. By definition $e=\frac{c}{a}=0.25$, so that $a=4c$.
Step 3. Use the relation $a^2-c^2=b^2$, we have $16c^2-c^2=(10^{10})^2=10^{20}$, thus
$c=\sqrt {10^{20}/15}\approx2.582\times10^{9}$ km and $a=4c\approx1.033\times10^{10}$
Step 4. The distance between the Pluto and the Sun at perihelion is given by $d_p=a-c=3c=7.746\times10^{9}$km
Step 5. The distance between the Pluto and the Sun at aphelion is given by $d_a=a+c=5c=1.291\times10^{10}$km
