Answer
$r=\dfrac{12}{2+3 \sin \theta}$
and the conic is a hyperbola as $e \gt 1$
Work Step by Step
From the given statement , we have: Directrix , $a=4$ above the pole and eccentricity ; $e=\dfrac{3}{2}$
The equation must be of the form as follows: $r=\dfrac{ae}{1+e \sin \theta}$
Then, we have $r=\dfrac{(4)(3/2)}{1+(3/2) \sin \theta}$
Therefore, $r=\dfrac{12}{2+3 \sin \theta}$
and the conic is a hyperbola as $e \gt 1$
