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Loci Conics Parabola eqn in standard form

  1. Aug 24, 2005 #1
    I need to find the equation of a parabola in standard form if the domain of an arch under a bridge is {-50 <= x <= 50} and the range is {0<=y<=20}

    The span of the arch is 100 metres and the height is 20 metres.

    standard form of a parabola is

    [tex] (x-h)^2=4p(y-k) [/tex]

    i know the vertex is (0,20) so the equation should look like [tex] x^2=4p (y-20) [/tex] so far but im not sure what the focus is or how to determine it plz help me out
     
  2. jcsd
  3. Aug 24, 2005 #2

    Hurkyl

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    Do you know of any other points on your parabola?
     
  4. Aug 25, 2005 #3

    HallsofIvy

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    What are the coordinates of the ends of the bridge?
     
  5. Aug 25, 2005 #4
    Well yes I think I do know two more points

    (0,-50) and (0,50) what do I do to complete my standard form equation?

    Do I sub in one of the points into [tex] x^2=4p (y-20) [/tex] and then solve for p?

    I dont think i'm doing this right
    [tex] x^2=4p (y-20) [/tex]
    [tex] 0^2=4p(50-20) [/tex]
    [tex] 0=4p(30) [/tex]
    [tex] 0=120p [/tex]

    now what?? I dont know what to do please help me!
     
    Last edited: Aug 25, 2005
  6. Aug 25, 2005 #5

    HallsofIvy

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    Okay, if (0,50) and (0,-50) have to satisfy the equation, how about you put x= 0, y= 50 into the equation and see what happens?
     
  7. Aug 25, 2005 #6
    that's what I have done in the last post....I'm still not sure what to do? plzzzz help!!
     
  8. Aug 25, 2005 #7

    EnumaElish

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    Did you mistype "locus" as "focus"?

    Let's see: (0, 20) is the vertex. (-50, 0) and (50, 0) are the endpoints. If I understand what the span is, then the endpoints should have y = 0 instead of x = 0. (That is, I believe that you interchanged the x value and the y value of the two endpoints by mistake.)

    You have [tex](x-h)^2=4p(y-k)[/tex] with 3 parameters to calculate: h, k and p.

    You have the 3 points and should be able to solve the following 3 equations for h, k and p:

    [tex] (0-h)^2=4p(20-k) \ ... \text{ Eq. (1)}[/tex]
    [tex] (-50-h)^2=4p(0-k) \ ... \text{ Eq. (2)}[/tex]
    [tex] (50-h)^2=4p(0-k) \ ... \text{ Eq. (3)}[/tex]

    Let {h*,k*,p*} be the solution to (1)-(3). Then, the locus of points on the parabola is "the set of all (x,y) in [itex]\mathbb R^2[/itex] that satisfy the standard form equation when {h*,k*,p*} are substituted for {h, k, p} in that equation."
     
  9. Aug 26, 2005 #8
    Nope i meant focus not locus p stands for the focus of a parabola i dont understand what to do with all those equations. Can someone tell me how to solve for p?
     
  10. Aug 26, 2005 #9
    you know points (50,0) and (-50,0). solve equation [tex]x^2=4p (y-20)[/tex] again.
     
  11. Aug 26, 2005 #10

    EnumaElish

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    Eq. (1) is how [itex](x-h)^2=4p(y-k)[/itex] looks when x = 0, y = 20.
    Eq. (2) is how [itex](x-h)^2=4p(y-k)[/itex] looks when x = -50, y = 0.
    Eq. (3) is how [itex](x-h)^2=4p(y-k)[/itex] looks when x = 50, y = 0.
     
  12. Aug 27, 2005 #11
    I'm sorry I see all the equations but they are all the same and I still dont know how to solve for p.
     
  13. Aug 27, 2005 #12

    HallsofIvy

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    No, they are NOT the same! When EnumaElish said
    "Eq. (1) is how [itex](x-h)^2= 4p(y-k)[/itex] looks when x = 0, y = 20."

    He really expected you to DO that: put x= 0, y= 20 and see what you get. Putting x=0, y= 20, then x= -50, y= 0, then x= 50, y= 0 into
    [itex](x-h)^2= 4p(y-k)[/itex] gives you three different equations that you can then solve for h, p, and k. (in fact, you could probably guess h and k!)
     
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