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ckallerid
 one year ago
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
ckallerid
 one year ago
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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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0What is the directrix? How can we use this to help us derive the parabola?

ckallerid
 one year ago
Best ResponseYou've already chosen the best response.0Honestly I have no clue, Can you give me like a quick run down on how to do any of this?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The 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 \[ (xh)^2 + (yk)^2 = (yh_1)^2\], 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Solve for y in this equation and you will get the equation of your parabola.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you can answer this in like two seconds if you know two things first, what does this look like?

ckallerid
 one year ago
Best ResponseYou've already chosen the best response.0a simple plug in problem.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1443317214435:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0there is a crude picture with a focus and directrix labelled do you know what the parabola looks like?

ckallerid
 one year ago
Best ResponseYou've already chosen the best response.0I'm guessing it would look vertical??

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0copy my picture and draw it

ckallerid
 one year ago
Best ResponseYou've already chosen the best response.0dw:1443317495455:dw(like I said I have no clue)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the vertex is half way between \((0,1)\) and \(y=1\) like this

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1443317677014:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the 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
 one year ago
Best ResponseYou've already chosen the best response.0that 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
 one year ago
Best ResponseYou've already chosen the best response.0your answer is therefore \[4y=x^2\]

ckallerid
 one year ago
Best ResponseYou've already chosen the best response.0Didn't I need to solve for Y or something like that?

ckallerid
 one year ago
Best ResponseYou've already chosen the best response.0whatevs thanks guys!
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