Chapter 10 - Analysis of Variance - Chapter 10 Test Prep - Review Exercises - Page 1000: 10

$${x^2} = - 12y$$

$$\eqalign{ & {\text{focus}}\left( {0, - 3} \right) \cr & {\text{The focus is }}\left( {0,p} \right),{\text{ then the equation is of the form }} \cr & {x^2} = 4py \cr & {\text{with }}p = - 3 \cr & {x^2} = 4\left( { - 3} \right)y \cr & {x^2} = - 12y \cr} $$