anonymous
  • anonymous
Derive the equation of the parabola with a focus at (0, 1) and a directrix of y = −1.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
@zepdrix
zepdrix
  • zepdrix
eyyy
zepdrix
  • zepdrix
I think I remember you asking this yesterday? :O

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More answers

zepdrix
  • zepdrix
|dw:1438975490650:dw|
zepdrix
  • zepdrix
Our Vertex is always equidistant from our focus and directrix, ya? So that puts us at 0,0
zepdrix
  • zepdrix
|dw:1438975604620:dw|
zepdrix
  • zepdrix
Umm from here... we can either do it the long way... or use that nice shortcut formula.. im trying to remember what it looks like..
zepdrix
  • zepdrix
Here is that shortcut that I was thinking of: \[\large\rm (x - h)^2 = 4p (y - k)\]Where our vertex is the point (h,k), and p is the distance between the focus and vertex.
anonymous
  • anonymous
yes thank you i did ask yesterday but i had trouble at the end and couldnt find a :(
anonymous
  • anonymous
h and k i got as 0, 0
zepdrix
  • zepdrix
\[\large\rm (x-0)^2=4p(y-0)\]\[\large\rm x^2=4py\]Ok good. let's solve for y,\[\large\rm y=\frac{1}{4p}x^2\]
zepdrix
  • zepdrix
And then, what is the value of p? The distance from the vertex up to the focus?
anonymous
  • anonymous
wouldnt it be 2? from 1 to -1? or no
zepdrix
  • zepdrix
|dw:1438975963650:dw|No that would be 2p :)
anonymous
  • anonymous
|dw:1438976050358:dw|is this the focus? at 0,1
zepdrix
  • zepdrix
|dw:1438976120582:dw|yes :) i want to know the distance from here to here from the VERTEX to the FOCUS. not from the DIRECTRIX silly :)
anonymous
  • anonymous
ohhh 1
anonymous
  • anonymous
sorry sorry
zepdrix
  • zepdrix
\[\large\rm y=\frac{1}{4p}x^2 \qquad\to\qquad y=\frac{1}{4\cdot1}x^2\]
zepdrix
  • zepdrix
Yayyy team, we did it! \c:/
anonymous
  • anonymous
okay i got it now thank you soo much
anonymous
  • anonymous
:))

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