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Topic review (newest first)

John E. Franklin
2006-01-10 14:56:42

Hi Justin.  This may not help, but who knows?

Define the bottom or top of the parabola at (d,D)

I'll rename your (x1,y1) to be (a,A), just because I feel like it.
And (x2,y2) is (b,B).  And (x3,y3) is (c,C).
Also your "a" multiplier, I will call "N", as it makes the parabola narrower.

Now if you get the point (d,D) and find N, then the equation for the parabola will be
(y-D) = N (x-d)^2

Okay, let's begin:

There are distance relationships in a parabola to do with squares,
hence we can write these equations:

N(a-d)^2 = |A-D|
N(b-d)^2 = |B-D|
N(c-d)^2 = |C-D|

2006-01-10 05:53:35

I am trying to write code to construct a parabola based on three points which are randomly determined at run-time. The problem I am having is that I cannot reduce any equations due to the fact that none of the variables have known values at the time of writing the equations (they will only be known at run-time).

This is as far as I have gotten. I defined c first because it seemed the easiest to isolate. I went for a next, but, as you can see, I ran into a problem where a is defined partially in terms of itself. I am looking to have one of the three (a, b, or c) defined only in terms of the X/Y coordinates. It doesn't matter how long or complicated the equation is. Then one of the other two will be defined in terms of the X/Y cooordinates as well as that previously defined variable. Then the final one will be defined according to anything as long as it doesn't have itself in the definition.

y1 = a*x1^2 + b*x1 + c
y2 = a*x2^2 + b*x2 + c
y3 = a*x3^2 + b*x3 + c

c = y1 - (a*x1^2 + b*x1)

y2 = a*x2^2 + b*x2 + y1 - (a*x1^2 + x1*b)
a*x2^2 = b*x2+y1 - (a*x1^2 - x1*b) - y2
a = (b*x2 + y1 - (a*x1^2 - x1*b) - y2) / x2^2

Anyone able to help with this? Thanks.

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