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Math Help - Parabola help please

  1. #1
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    Parabola help please

    I'm not sure if this falls under calculus as its been a while since I took it.

    I'm writing a program and I need help with finding projectile parabola equations.

    Rather than explain the whole situation I drew a quick, but poor, picture in paint.net.



    Basically I need know, that given x1,y1,x2,y2, and y0, can I find x0.

    I already solved another situation where x0,y0 is known but another point is missing. I just cannot seem to do it when the center point x-value is not known.

    It seems like it should be simple, but maybe I'm just tired. I could use some help please.
    Attached Thumbnails Attached Thumbnails Parabola help please-parab.png  
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  2. #2
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    A parabola has a general equation

     (x - x_0)^2 = 4a(y-y_0)

    If  x_0 is unknown, simply substitute  (x_1,y_1) and (x_2,y_2) into the equation separately.

    This gives:
     (x_1 - x_0)^2 = 4a(y_1-y_0)
     (x_2 - x_0)^2 = 4a(y_2-y_0)

    Dividing one by the other:

     \left(\frac{x_1 - x_0}{x_2 - x_0}\right)^2 = \left(\frac{y_1 - y_0}{y_2 - y_0}\right)

    Since we know the values  y_1, y_2, and  y_0 , \frac{y_1 - y_0}{y_2 - y_0} is a constant. Lets call it  C^2 for convenience's sake.
     \left(\frac{x_1 - x_0}{x_2 - x_0}\right)^2 = C^2

     x_1 - x_0 = (x_2 - x_0) C = x_2C - x_0 C

     x_0 C - x_0 = x_2C - x_1

     x_0 (C - 1) = x_2C - x_1

     x_0 = \frac{x_2C - x_1}{C-1}

    We know that  x_1, x_2 and  C are constants, so you'll be able to solve for  x_0

    Overall, the equation should be:

     x_0 = \frac{x_2 \sqrt{\frac{y_1 - y_0}{y_2 - y_0}} - x_1}{\sqrt {\frac{y_1 - y_0}{y_2 - y_0}} - 1 }

    You may want to check my algebra. If you have any questions please ask.
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  3. #3
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    I appreciate the response but it does not seem to work. First I don't know where you got that general form because I was always taught the general form of a parabola is y=ax2+bx+c, and the standard form y=a(x-h)2+k.

    I tested a simple case anyway.
    Imagine...
    (x1,y1) = (0,0)
    (x2,y2) = (6,1)
    (x0,y0) = (x,4)
    obviously it is a downward pointing parabola and the missing x value should lie somewhere between 0 and 6, but most likely close to the midpoint since the y values of the side points only differ by 1.

    I plug in the (y1-y0)/(y2-y0) = (0-4)/(1-4) = 4/3
    so plugging in the rest...
    6 sqrt(4/3) / (sqrt(4/3) - 1) = 6.9282/.1547 = 44.78 which is way off.
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  4. #4
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    I think we may have a slight misunderstanding. I assumed  (x_0,y_0) is the vertex of the parabola. Is this so? If not my previous post is irrelevant. You cannot find an equation for a parabola, and consequently  x_0 with just two points.

    Quote Originally Posted by gyan1010 View Post
    I appreciate the response but it does not seem to work. First I don't know where you got that general form because I was always taught the general form of a parabola is y=ax2+bx+c, and the standard form y=a(x-h)2+k.
     (x - x_0)^2 = 4a(y-y_0)
     y - y_0 = \frac{1}{4a} (x - x_0)^2
     y = \frac{1}{4a} (x - x_0)^2 + y_0 Which is in your from y =a(x-h)2+k.

    I plug in the (y1-y0)/(y2-y0) = (0-4)/(1-4) = 4/3
    so plugging in the rest...
    6 sqrt(4/3) / (sqrt(4/3) - 1) = 6.9282/.1547 = 44.78 which is way off.
    Note that  C = \pm \sqrt {\frac{y_1 - y_0}{y_2 - y_0}} . Given two points, you have no way of knowing which one is pointing up or down. Substitute  C = -\sqrt {\frac{y_1 - y_0}{y_2 - y_0}} and you should be getting a value of  x_0 for a downward pointing parabola. Without the negative it's an upward pointing parabola.
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  5. #5
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    My apologies, must have been tired because I didnt consider the negative square root. Since all of my parabolas are downward it was always wrong until I used the negative. Thanks so much for your help.
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