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Parabola conical whatever equations

  1. Jun 5, 2006 #1
    I am dazed and confused how does 2x^2=y end up being directrix y=-1/8?
     
  2. jcsd
  3. Jun 5, 2006 #2
    I forgot the steps on how to do this.
     
  4. Jun 5, 2006 #3

    Andrew Mason

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    Putting the parabola into form [itex]x^2 = 4py[/itex] where p is the focus, will give you the location of the focus, p. Since the points on the parabola are equidistant form the directrix and focus, that should enable you to find the equation of that line.

    AM
     
  5. Jun 5, 2006 #4
    I don't see the math in it though...
    [itex]x^2 = 4py[/itex]
    [itex]x/4^2 = py[/itex]
    making [itex]x/4^2 = p[/itex]
    is it something like that?
     
  6. Jun 6, 2006 #5

    Andrew Mason

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    No. [itex]2x^2 = y[/itex] so [itex]x^2 = \frac{1}{2}y[/itex]. In the form [itex]x^2 = 4py[/itex] 4p = 1/2 and p = 1/8. So the focus is (0,1/8). Since the vertex is (0,0) which is equidistant from the focus and the directrix, the directrix is the horizontal line passing through (0,-1/8) ie. y = -1/8

    AM
     
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