Answer
$r=\dfrac{40}{4-5 \sin \theta}$
and the conic is a hyperbola as $e \gt 1$
Work Step by Step
From the given statement , we have: Directrix , $a=8$ below the pole and eccentricity ; $e=\dfrac{5}{4}$
The equation must be of the form as follows: $r=\dfrac{ae}{1-e \sin \theta}$
Then, we have $r=\dfrac{(8)(5/4)}{1-(5/4) \sin \theta}$
Therefore, $r=\dfrac{40}{4-5 \sin \theta}$
and the conic is a hyperbola as $e \gt 1$
