## Precalculus (6th Edition) Blitzer

a. $x^2=3y$ b. $0.75\ in$ above the vertex.
a. Assume the parabola equation is $x^2=4py$. Since point $(3,3)$ is on the curve, we have $3^2=4p(3)$, which gives $p=\frac{3}{4}=0.75$ and the equation $x^2=3y$ b. The focus is at $(0,p)$ or $(0,0.75)$; that is, $0.75\ in$ above the vertex.