Find the standard form of the equation of the parabola with a focus at (3, 0) and a directrix at x = -3.
a) y = (1/12)x^2
b) -12y = x^2
c) x = (1/12)y^2
d) y^2 = 6x

Notice the picture below

the directrix is a vertical line, and the focus point is to the right of it
the vertex is always half-way between those two, at h,k coordinates

thus the parabola is opening sideways, to the right, thus the "y" is the
variable that's at the second power

or 

so... use those values, and plug them in the equation

Hope this ; )

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