Find the vertex, focus, directrix, and focal width of the parabola. x2 = 12y

[tex]\bf \textit{parabola vertex form with focus point distance}\\\\ \begin{array}{llll} 4p(x- h)=(y- k)^2 \\\\ \boxed{4p(y- k)=(x- h)^2} \end{array} \qquad \begin{array}{llll} vertex\ ( h, k)\\\\ p=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix} \end{array}\\\\ -------------------------------\\\\ x^2=12y\implies (x-0)^2=12(y-0)\implies (x-\stackrel{h}{0})^2=4(\stackrel{p}{3})(y-\stackrel{k}{0})[/tex]

so, the vertex of h,k is at the origin, as you can see 0,0.

4p is the focal width, which is well, 4*3.

now, the parabola has a positive leading term's coefficient, namely 1x², the 1 is positive, well, that simply means that, since the squared variable is the "x", is a vertical parabola, and since its coefficient is positive, is opening upwards, like a bowl.

 the focus point will be "p" distance from the vertex, and the directrix is also "p" distance from the vertex as well, but in the opposite direction.

since the parabola is opening upwards, that's where the focus point is at, 3 units UP from 0,0, namely at (0,3).

and due to the same reason as above, the directrix is 3 units DOWN from the vertex, namely a horizontal line running over y = -3.