How do I find the directrix and focal length of a parabola? For example, y=1/12x^2

Question

stamp · Answer

any point on a parabola is equidistant to the focus and the directrix

anonymous · Answer

But how do I find the directrix...?

stamp · Answer

there is probably a formula

anonymous · Answer

Ok, thanks.

amistre64 · Answer

i believe the form is properly:  4py =x^2

anonymous · Answer

Oh, it is D:?

amistre64 · Answer

y = 4x^2 has a focal point at (0, 1/16)

if we assume y=4px^2, that would make the focal point at (0,1) ...

anonymous · Answer

Hm...that does seem to make sense. I put an example in my question. How would that work out?

amistre64 · Answer

y=1/12x^2  ; *12

12y = x^2

let 4p = 12 ,  p = 3  gives the distance we need from the vertex to find the focus and the directrix

amistre64 · Answer

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