Answer
$${y^2} - 4y - 12x + 4 = 0$$
Work Step by Step
$$\eqalign{
& {\text{Vertex: }}\left( {0,2} \right) \cr
& {\text{Directrix: }}x = - 3 \cr
& {\text{The directrix is }}x = - p + h{\text{ and the vertex is }}\left( {h,k} \right),{\text{so the}} \cr
& {\text{equation of the parabola is of the form}} \cr
& {\left( {y - k} \right)^2} = 4p\left( {x - h} \right) \cr
& {\text{Vertex}}\left( {h,k} \right):\left( {0,2} \right) \to h = 0,{\text{ }}k = 2 \cr
& x = - p + h = - 3 \cr
& - p + 0 = - 3 \cr
& p = 3 \cr
& \underbrace {{{\left( {y - k} \right)}^2} = 4p\left( {x - h} \right)}_ \downarrow \cr
& {\left( {y - 2} \right)^2} = 4\left( 3 \right)\left( {x - 0} \right) \cr
& {\left( {y - 2} \right)^2} = 12x \cr
& {\text{Expand}} \cr
& {y^2} - 4y + 4 = 12x \cr
& {y^2} - 4y - 12x + 4 = 0 \cr} $$
