To analyze the given polar equation \( r = \frac{4}{2 + 3 \sin \theta} \), we can convert it into a more recognizable form.
This equation resembles the general form of a conic section in polar coordinates:
\[ r = \frac{ed}{1 + e \sin \theta} \]
where \( e \) is the eccentricity and \( d \) is the distance from the pole to the directrix.
In our case, we need to manipulate the given equation to match this form:
\[ r = \frac{4}{2 + 3 \sin \theta} = \frac{4}{2} \cdot \frac{1}{1 + \frac{3}{2} \sin \theta} = 2 \cdot \frac{1}{1 + \frac{3}{2} \sin \theta} \]
This shows that:
- \( e = \frac{3}{2} \) (the coefficient of \( \sin \theta \))
- \( ed = 2 \) (from the \( r = \frac{ed}{1 + e \sin \theta} \) form)
Thus, we can find \( d \):
\[ d = \frac{ed}{e} = \frac{2}{\frac{3}{2}} = \frac{4}{3} \]
Since \( e = \frac{3}{2} > 1 \), this indicates that the curve represented by this equation is a hyperbola.
Moreover, the directrix is vertical since the conic form has \( \sin \theta \). Given the equation, we identify the directrix based on \( d \):
- The horizontal directrix (due to \( \sin \theta \)) will be at a distance of \( \frac{4}{3} \) units above the pole.
Therefore, the correct description of the graph represented by the equation is:
hyperbola, horizontal directrix at a distance of \( \frac{4}{3} \) units above the pole.