Answer
a. $r= \frac{2}{1+2sin\theta}$ .
b. $e=2$, $p=1$, hyperbola.
c. See figure.
Work Step by Step
a. Rewrite the equation $r=\frac{6}{3+6sin\theta}=\frac{2}{1+2sin\theta}$ as the standard form of a conic in polar coordinates.
b. From the above equation, we can determine $e=2$ and $ep=2$, which gives $p=1$. With $e\gt1$, the equation can be identified as a hyperbola.
c. We can graph the polar equation as shown in the figure.
