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I have the X,Y of 2 end points and 1 bezier point, of a Quadratic Bezier curve.

Using this data, how can I derive the polynomial representation of the curve?

alt text
(source: euclidraw.com)

Glorfindel
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Robin Rodricks
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3 Answers3

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B(t) = (1-t) * (1-t) * B0 + 2 * (1-t) * t * B1 + t * t * B2

joel.neely
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  • Oh. If that's what the OP was looking for, then you're exactly correct. (at least Wikipedia corroborates it ;) I thought he was trying to get a polynomial equation without the parameter t in it. – Jason S Feb 21 '09 at 18:48
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Oog. That would be tricky. Beziers are parametrized curves, namely:

x = f(t)
y = g(t)

where t=0 yields one endpoint and t=1 yields the other.

You could technically figure out how to eliminate "t" and get an equation in x and y, but it would not be a polynomial like y = a + bx + cx2 ...; it would be an equation h(x,y) = 0 where h is probably somewhat ugly.

Jason S
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Wikipedia has a section about this. Perhaps this helps.

schnaader
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