anonymous
  • anonymous
Find the standard form of the equation of the parabola with a focus at \((0,−9)\) and a directrix \(y=9.\)
Mathematics
schrodinger
  • schrodinger
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welshfella
  • welshfella
this parabola opens upwards and standard form is x^2 = 4ay where a is the y coordinate of the focus and y = -a is the directrix so can you work this one out?
anonymous
  • anonymous
Wouldn't the parabola open downwards?
anonymous
  • anonymous
It would. The form for a parabola is \[y = 1/4p(x-h)^2+k\] From here you can plug 9 into p since p = distance between vertex and focus. The h and k values are both 0 since the vertex is at (0,0). Can you get it from here?

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anonymous
  • anonymous
I lied, plug -9 in for p.
anonymous
  • anonymous
|dw:1439345400610:dw|
anonymous
  • anonymous
P is (x,y)
anonymous
  • anonymous
Hang on, I'm trying to solve it. :)
anonymous
  • anonymous
Okay, post your answer here so I can check it for you :)
anonymous
  • anonymous
So is it \(y=-\frac{ 1 }{ 36 }x^2?\)
anonymous
  • anonymous
It sure is!
anonymous
  • anonymous
Wow thanks! I was working with this, but I got the vertex wrong. :( How would you find the vertex?
anonymous
  • anonymous
No problem! Okay so the vertex is equidistant from both the focus and the directrix. Since they both are -9 away from the origin (as the x value for both is 0) you can tell that the vertex must be at the origin in order for it to be equidistant from both!
anonymous
  • anonymous
Awesome! Thanks again. You're great! :)
welshfella
  • welshfella
yes surit is correct - i misread the problem
welshfella
  • welshfella
- and migillope also

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