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Directrix and Focus

  1. Aug 13, 2005 #1
    vertex (-2, 3), focus (0,3)

    Find the equation of the parabola.

    Dirac.
     
  2. jcsd
  3. Aug 13, 2005 #2

    lurflurf

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    What have you tried so far?
    Find the directix. then recall
    A parabola is the set of point equidistant from the focus and directix.
     
    Last edited: Aug 13, 2005
  4. Aug 13, 2005 #3
    So far I have got to this

    ((y-3)^2)=4a(x+2)

    What now

    Dirac.
     
  5. Aug 13, 2005 #4

    lurflurf

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    a will be the distance between the focus and vertex.
    Why is this?
     
  6. Aug 13, 2005 #5
    How does one find the vertex
     
  7. Aug 13, 2005 #6

    lurflurf

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    you started out with
    vertex (-2, 3), focus (0,3)
    The points of a parabola are equidistant from it directrix and focus.
    The vertex is the point closest to them. that is
    distance(vertex,focus)<=distance(point on parabola,focus)
    with equality only when the point is the vertex.
     
  8. Aug 13, 2005 #7
    Could you just post the answer, or would it get up your backside to post something useful in your life

    Dirac.
     
  9. Aug 13, 2005 #8

    lurflurf

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    I already said you a is the distance between
    vertex (-2, 3), focus (0,3)
    here you go
    a=sqrt[(0-(-2))^2+(3-3)]
    =sqrt[2^2]
    a=2
    Do you understand why?
    Can you find the equation of a parabola with
    vertex (0,0), focus (0,-4)
     
  10. Aug 13, 2005 #9

    HallsofIvy

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    How pathetic. Why in the world would anyone care what the answer to some made up question like this is? The whole point of an exercise like this is to learn how to get the answer. I'll bet there's a "general formula" for the parabola in the same section of your text as this problem. Look it up. We're here for people who want to learn not for people who want the answers handed to them.
     
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