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- ckallerid

I require assisstance!
Derive the equation of the parabola with a focus at (0, 1) and a directrix of y = −1.
f(x) = −one fourth x2
f(x) = one fourth x2
f(x) = −4x2
f(x) = 4x2

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- ckallerid

- katieb

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- anonymous

What is the directrix? How can we use this to help us derive the parabola?

- ckallerid

Honestly I have no clue, Can you give me like a quick run down on how to do any of this?

- anonymous

The directrix is a special line to the parabola. Consider some point on the parabola, then the distance from the diretrix to this point and the distance from the focus to this point are the same. Mathematically, we can write this as
\[ (x-h)^2 + (y-k)^2 = (y-h_1)^2\], where (h,k) is the focus of the parabola and y = h_1 is the equation of the directrix. In this case, h = 0, k = 1, and h_1 = -1.

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- anonymous

Solve for y in this equation and you will get the equation of your parabola.

- anonymous

you can answer this in like two seconds if you know two things
first, what does this look like?

- ckallerid

a simple plug in problem.

- anonymous

|dw:1443317214435:dw|

- anonymous

there is a crude picture with a focus and directrix labelled
do you know what the parabola looks like?

- ckallerid

I'm guessing it would look vertical??

- anonymous

copy my picture and draw it

- ckallerid

|dw:1443317495455:dw|(like I said I have no clue)

- anonymous

yeah i can see that

- anonymous

the vertex is half way between \((0,1)\) and \(y=-1\) like this

- anonymous

|dw:1443317677014:dw|

- anonymous

the point half way between \((0,1)\) and \(y=-1\) is \((0,0)\) so the vertex is at the origin, and the parabola opens up

- anonymous

that means it looks like \[4py=x^2\] where \(p\) is the distance between the vertex and the focus, or the vertex and the directrix
in this case \(p=1\)

- anonymous

your answer is therefore \[4y=x^2\]

- ckallerid

Didn't I need to solve for Y or something like that?

- ckallerid

whatevs thanks guys!

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