Hello, Gracy!
You're expected to know the general form for polar conics.
. . $\displaystyle r \:=\:\frac{ke}{1 + e\sin\theta}\qquad\text{or}\qquad r\:=\:\frac{ke}{1 + e\cos\theta}$ . where $\displaystyle e$ = eccentricity
And these rules: .$\displaystyle \begin{array}{ccc} e < 1 & \text{ellipse} \\ e = 1 & \text{parabola} \\ e > 1 & \text{hyperbola}\end{array}$
Given the polar equation: .$\displaystyle r\:=\:\frac{4}{2+2\sin x}$
Find the directrix, eccentricity and sketch the conic.
We have: .$\displaystyle r \:=\:\frac{2}{1 + \sin x}\quad\Rightarrow\quad\boxed{e = 1}$
Since $\displaystyle e = 1$, we have a parabola with its focus is at the pole (origin).
Plot a few points: .$\displaystyle \begin{array}{ccc}x = 0 & r = 2 \\ x = \frac{\pi}{2} & r = 1 \\ x = \pi & r = 2\end{array}$
The graph looks like this: Code:
2
      +      

1
***
*  *
*  *
*+*

*  *

The focus is 1 unit from the vertex.
. . Hence, the directrix is also 1 unit from the vertex.
It is the horizontal line: $\displaystyle y \:=\:2$
. . Converting to polars: .$\displaystyle r\sin x = 2\quad\Rightarrow\quad\boxed{ r = 2\csc x}$