Answer
(a) e = 1
(b) Parabola
(c) y = 2/3
(d) Image below:
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Work Step by Step
(a)
$$r = \frac {2}{3 + 3sin(\theta)} \div \frac{3}{3} = \frac {2/3}{1 +sin(\theta)}$$
Based on the original equation:
$$\frac{ed}{1 + esin(\theta)}$$
$$e = 1$$
(b)
$e = 1 $, therefore, this equation represents a parabola.
(c)
The equation has a $+sin(\theta)$ and the focus is at (0,0), thus, the equation of the directrix must be in the "$y = c$" pattern.
In this case: $c = d$:
$$ed = 2/3 \longrightarrow d = \frac{2/3}{e} = \frac{2/3}{1} = 2/3 $$
$$y = 2/3$$
(d)
Plot the points where $\theta = 0$, $\theta = \pi/2$, $\theta = \pi$
$$r = \frac {2}{3 + 3sin(0)} \longrightarrow (2/3, 0)$$
$$r = \frac {2}{3 + 3sin(\pi/2)} \longrightarrow (1/3, \pi/2)$$
$$r = \frac {2}{3 + 3sin(\pi)} \longrightarrow (2/3, \pi)$$
Draw a parabola that passes through these points.
