\begin{aligned} & \text{1. Determine which form of the parabola equation to use. Since the vertex and focus differ in the y-coordinate only, } \\ & \text{the parabola opens vertically \lparen upwards or downwards\rparen. We use the standard form }(x-h)^2=4p(y-k) \\ & \text{2. Identify the vertex }(h,k)\text{ and the focus }(h,k+p). \\ & \text{3. From the given data: }(h,k)=(-5,-6)\text{ and focus}=(-5,-3) \\ & \text{4. Calculate }p \\ & k+p=-3 \\ & -6+p=-3 \\ & p=3\text{ \lparen focus is 3 units above the vertex, so p is negative since the parabola opens downwards\rparen} \\ & \text{4. Substitute the values into the standard form }(x-h)^2=4p(y-k) \\ & (x-(-5))^2=4(3)(y-(-6)) \\ & (x+5)^2=12(y+6) \\ & \text{The equation of the parabola is }(x+5)^2=12(y+6) \\ & \placeholder{}\end{aligned}