Answer
(a) See explanations.
(b) $r=\frac{1.49\times10^8}{1-0.017cos\theta}$
Work Step by Step
(a) Step 1. Write the general polar equation for an ellipse as $r=\frac{ed}{1-e\cdot cos\theta}$.
Step 2. Use the relation $a^2=\frac{e^2d^2}{(1-e^2)^2}$ given on page 825, we have $d^2=\frac{a^2(1-e^2)^2}{e^2}$ and $d=\frac{a(1-e^2)}{e}$ (here $e\lt1$ and we take $d$ as positive)
Step 3. Use the expression of $d$ in the equation of step-1, we have $r=\frac{e\times\frac{a(1-e^2)}{e}}{1-e\cdot cos\theta}=\frac{a(1-e^2)}{1-e\cdot cos\theta}$ which is the formula given in the Exercise.
(b) In this case, given $e=0.017$ and $2a=2.99\times10^8$km, use the formula in part (a), we have
$r=\frac{1.495\times10^8(1-0.017^2)}{1-0.017cos\theta}$ or $r=\frac{1.49\times10^8}{1-0.017cos\theta}$
