Answer
(a) $\frac{4}{5}$
(b) Ellipse
(c) $y=-1$
(d) See graph
Work Step by Step
Part (a)
$r=\frac{4}{5-4\sin\theta}$
$\frac{ed}{1-e\sin\theta}=\frac{\frac{4}{5}}{1-\frac{4}{5}\sin\theta}$
$e=\frac{4}{5}$ and $ed=\frac{4}{5}$
$e=\frac{4}{5}$ and $d=1$
So, the eccentricity is $\frac{4}{5}$.
Part (b)
Since $e=\frac{4}{5}<1$, the conic is an ellipse.
Part (c)
We have obtained $d=1$.
Using Theorem 6, and part (d) of Figure 2, the directrix has the equation $y=-1$
Part (d) Graph the conic.
