Answer
(a) $e=\frac{3}{5}$, directrix $y=-\frac{2}{3}$.
(b) $r=\frac{1}{5-3sin(\theta-\frac{2\pi}{3})}$
(c) See graph.
Work Step by Step
The standard form of a conic polar equation is $r=\frac{ed}{1\pm e\cdot cos\theta}$ or $r=\frac{ed}{1\pm e\cdot sin\theta}$. The graph of the equation will be a parabola if $e=1$, an ellipse if $e\lt1$, and a hyperbola if $e\gt1$
(a) Rewrite the equation given in the Exercise as $r=\frac{2/5}{1-\frac{3}{5}sin\theta}$, we can identify that $e=\frac{3}{5}\lt1$ by comparing it with a standard equation above. Thus the graph of the equation will be an ellipse. With $ed=\frac{2}{5}$, we get $d=\frac{2}{3}$ and the directrix is given by $y=-d=-\frac{2}{3}$.
(b) The equation after the rotation can be written as $r=\frac{1}{5-3sin(\theta-\frac{2\pi}{3})}$
(c) See graph.
