Answer
$(x-\frac{1}{2})^2=2(y-1)$
vertex $V(\frac{1}{2}, 1)$, focus$F(\frac{1}{2}, \frac{3}{2})$, directrix $y=\frac{1}{2}$
See graph.
Work Step by Step
Step 1. Form squares with the variables: $4(x^2-x+\frac{1}{4})=8y-9+1$ or $(x-\frac{1}{2})^2=2(y-1)$
Step 2. The above equation represents a parabola with vertex at $(\frac{1}{2}, 1)$
Step 3. With $4p=2$, we have $p=\frac{1}{2}$, the focus is at $(\frac{1}{2}, \frac{3}{2})$
Step 4. The directrix can be found as $y=1-\frac{1}{2}=\frac{1}{2}$
Step 5. See graph.
