For an equation of the form y = a(x - b)^2 + c, where a, b, c are all real numbers, the vertex is at (b , c).
In this case, we have y = (1/12)(x - 0)^2 + 0, thus the vertex is (0 , 0).

Rearranging y = (1/12)x^2, we have x^2 = 12y

Suppose we have an equation of the form (x - a)^2 = b(y - c), where a, b, c are all real numbers.
Suppose the vertex is (d , e), where d, e are some real numbers.
The focus will be (d , e + (b/4))
The directrix will be y = e - (b/4)

Notice that in our case, b = 12
Thus, b / 4 = 12 / 4 = 3
Our vertex is (0 , 0).
Hence, the focus is (0 , 0 + 3) = (0 , 3).