Answer
focus $(0,-\frac{E}{4A})$, directrix $y=\frac{E}{4A}$
Work Step by Step
Step 1. Rewriting the equation as $x^2=-\frac{E}{A}y$, we have $4p=-\frac{E}{A}$ and $p=-\frac{E}{4A}$
Step 2. The vertex is at $(0,0)$. We can identify the focus at $(0,-\frac{E}{4A})$ with directrix $y=\frac{E}{4A}$
