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y = -1/4 x^2
y^2 = -4x
y^2 = -16x
y = -1/16 x^2

Just a second...

no problem

ok so what am i plugging in?

Oh, wrong question. You have the vertex and the focus, but not the directrix.

yea thats it

There's a process to find the directrix.

thats where im lost

Did you know that any point on the parabola is equidistant from the focus and the directrix?

This means that we can easily find the directrix since we have the vertex.

so how?

oh ok i understand. so then from there i would plug it in to the equation you gave earlier?

do i need to solve for the missing x in the directrix

You don't. You insert the x into the formula in place of \(x_2\)

oh ok

how would i do that? im on the computer and im doing my work on paper...

Use the draw button or use \(LaTeX\)

ok ill have to draw it because my computer doesn't support latex

You can still type the LaTeX. It won't stop it from showing up on my end.

just give me a couple minutes

There's a draw button you can click BTW.

i know im just in the middle of doing the work

|dw:1439499465843:dw|

ok ill start off with showing you what i plugged in:
(x-0)^2 + (y-(-4))^2 = (x-x)^2 + (y-4)^2

Looks good so far.

x^2 + y^2 + 8y + 16 = y^2 - 8y +16

Yes you can do it that way, but there's a way to do it that avoids expanding.

What do you get for your final simplified result?

give me a sec

Once you get the correct result, I'll show you the other way to do it.

y=x^2/16

my answer would be D

D is correct, but you forgot the negative in your answer.

yea i forgot to type it

So \(LaTeX\) does not load for you?

no.... :(

If not, I can do it by hand and upload it that way.

sure

Hang on, I'm about to upload it

ok

|dw:1439501181613:dw|

ok i see. i understand it. thanks for all the help.

You're most welcome :) You're a great student.