## Precalculus (6th Edition) Blitzer

Step 1. Using the given equation with parameter values, for Mercury, we have $r=\frac{(1-e^2)a}{1-e\ cos\theta}=\frac{(1-0.2056^2)(36.0\times10^6)}{1-0.2056\ cos\theta}$ Step 2. For Earth, we have $r=\frac{(1-0.0167^2)(92.96\times10^6)}{1-0.0167\ cos\theta}$ Step 3. Graph both equations (red for Mercury) as shown in the figure. Step 4. We can see that both orbits are close to circles and they should have a common focus (the Sun).