If my directrix is -1, and my vertex is 0,0, what formula is used to find the standard form of the equation?

Question

I know the answer; just need to see how they got it.

Answer 1

There are two that you could use. I will show you with the one I am most comfortable with:

y = a(x-h)^2+k

Where the vertex of this equation is (h,k) and the directrix is y = k - 1/(4*a).

So in this case, the vertex is (0,0), which means that h=0, and k=0. The directrix is -1 which means that:

-1 = k - 1/(4*a)

or

-1 = -1/(4*a)

since k=0.

Solve for a to get the first part of the equation:

-4*a = -1 => a=1/4

Now, plug it back into the original equation:

y = (1/4)(x-0)^2 + 0 =>

y = (1/4)x^2

And now, subtract the (1/4)x^2 and multiply by -4 to get:

x^2 - 4y = 0.