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Homework Help: How to find the equation of a parabola with only 3 pts.

  1. Jul 29, 2008 #1
    If you have 3 points of a parabola that are'nt the vertex and the only other information you recieve is that the A.O.S is parallel to the y axis how would you acurately find the equation? Everytime I try, I always end up with odd fractions......... the pts are (-3,-4),
    (-1,-1), (2,3)



    2. Relevant equations: general form: y=ax^2+ bx+c standard form: y-h=a(x-k)^2 a=1/4l l is the distance from the vertex to either the directrix or the focus, also that the equations opens either up or down....



    Attempt:(-3,-4) -4=9a-3b+c (-1,-1) -1=a-b+c (2,3) 3=4a+2b+c

    (-1)(-4=9a-3b-c) (-1)(-4=9a-3b-c)
    (-1= a -b+c) (3=4a+2b+c)
    + ___________ +____________
    3=-8a+2b 7=-5a+5b


    (-5)(3=-8a+2b)
    ( 2)(7=-5a+5b)
    +_____________
    1=30a a=1/30

    7=-5(1/30)+5b 7= -5/30+5b 215/30=5b 1075/30=b b= 215/6 a=1/30 b=215/6

    3=4a+2b+c 3=4/30+2(215/6)+c 86/30= 215/3+c c=-2064/30


    after that point: I thought something went wrong and attempted to redo it.........-.-
     
  2. jcsd
  3. Jul 29, 2008 #2

    berkeman

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    Staff: Mentor

    I get a = 1/30 also, using determinants to solve the equations for it. I think you're on the right track...
     
  4. Jul 29, 2008 #3

    berkeman

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    Staff: Mentor

    But in your solution for b, I think there's an issue...

    7=-5(1/30)+5b
    7= -5/30+5b
    215/30=5b
    1075/30=b <--- (see the issue?)
    b= 215/6
     
  5. Jul 30, 2008 #4
    okay thanks ^.^ I think I understand it more.
     
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