Answer
(a) $r=\frac{1}{1+0.5 cos\theta}$, see graph (left ellipse).
(b) Ellipse, see graph (right ellipse).
Work Step by Step
(a) Step 1. As $e=\frac{1}{2}$, the graph of the conic will be an ellipse.
Step 2. The directrix is vertical and on the right side of the focus (pole), we can assume a general form of the polar equation as $r=\frac{ed}{1+e\cdot cos\theta}$
Step 3. The equation of the directrix is $x=d$ from the above equation, thus we have $d=2$ based on the given condition.
Step 4. The polar equation for the ellipse is $r=\frac{\frac{1}{2}\times2}{1+\frac{1}{2} cos\theta}$ or $r=\frac{1}{1+\frac{1}{2} cos\theta}$
Step 5. See graph.
(b) Rewrite the equation as $r=\frac{3/2}{1-\frac{1}{2}sin\theta}$, we have $e=\frac{1}{2}$ so the equation represents an ellipse as shown in the graph.
