Question

Find the vertex, focus, directrix, and focal width of the parabola.

x = 3y2

Answer 1

The regular form for this problem would be:
\[(y-k)^2 = 4a(x-h) \]
To get the problem as close to that form as possible you would want to divide both sides by 3 so that you have:
\[y^2 = \frac{ 1 }{ 3 } x\]

The vertex would be (h,k).
The focus is (h+a,k)
The directrix is x=h-a
The focal is the 4a part.

To get a for the focus and the directrix you will divide the 1/3 by 4.

Let me know if you need further help. :)