Answer
parabola. vertex $V(0, 1)$, focus $F(0,-2)$, directrix $y=4$
See graph.
Work Step by Step
Step 1. Identify the type of the conics: rewrite the equation as $x^2=-12(y-1)$ the equation is of the form of a parabola.
Step 2. The vertex can be found at $V(0, 1)$
Step 3. With $4p=12$, we get $p=3$, so the focus is at $(0, 1-p)$ or $F(0,-2)$
Step 4. The directrix can be found as $y=1+p=4$
Step 5. See graph.
