Question

Find the standard form of the equation of the parabola with a focus at (3, 0) and a directrix at x = -3.

y = 1/12x^2

-12y = x^2

x = 1/12y^2

y^2 = 6x

Answer 1

notice that the focus is at (3, 0) and the directrix is at x = -3
thus

keep in mind that the focus is where the parabola opens at, at a certain distance from the vertex

and the directrix is EXACTLY the same distance from the vertex in the other way

so where do you think is our vertex ?

Answer 2

zero?

Answer 3

yes, the origin so... for a parabola of that type, horizontally opening, the "focus form" equation will be as \(\bf (y-k)^2=4p(x-h)\)

the vertex is at (h, k)
p = distance from the vertex to either the focus or the directrix

so just plug in your values

Answer 4

i don't get how

Answer 5

well... ok... what's the distance from the vertex to the say.. the focus? which is the same distance from the vertex to the directrix btw