Find the standard form of the equation of the parabola with a vertex at the origin and a focus at (0, -4).

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Find the standard form of the equation of the parabola with a vertex at the origin and a focus at (0, -4).

Mathematics
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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

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If given two points, the focus \((x_1, y_1)\) and the directrix \((x_2, y_2)\), you can insert them into the following formula: \((x - x_1)^2 + (y - y_1)^2 = (x - x_2)^2 + (y - y_2)^2\) then simplify afterwards to get the standard form of the parabola.
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?
Well, basically, we know that the focus is (0,-4) and the vertex is (0,0). Since the focus is 4 units below the vertex, that means the directrix is four units above it.
We also know that the directrix is a horizontal line. In this case it will be y = 4. When we express the directrix as a point, it becomes \((x, 4)\)
oh ok i understand. so then from there i would plug it in to the equation you gave earlier?
So now we have the focus (0,-4) and the directrix (x,4). Plug those points into the formula above to find the standard form of the equation of the parabola. Yes.
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
At this point, it's probably a good idea to show the work you've done so far that way I can make sure you've done this step correctly.
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|
You should post what you've done now, that way you don't get too far ahead because one wrong mistake and then you'll have so much re-work to do.
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.

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