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I would like to find a cubic Bézier curve x(t) knowing that x(t):[0,1]->R². I only have the two following points : x(0) = (0,0); x(1) = (9,0).

Also, I know that the curve intersects itself orthogonally at x(1/4) and x(3/4). I think that this condition can allow me to find the two missing control points, but I don't know how to use this information.

Once I have the 4 control points, I just have to compute the sum formula using Berstein basis, and this will give me the function x(t) that I'm looking for.

Thanks for your help!

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For a cubic Bezier curve $x(t)=P_0(1-t)^3+3P_1t(1-t)^2+3P_2t^2(1-t)+P_3t^3$, the $x(0)=(0,0)$ and $x(1)=(9,0)$ give

$P_0=(0,0)$
$P_3=(9,0)$

The curve intersects orthogonally at $x(1/4)$ and $x(3/4)$ means

$x(1/4) = x(3/4)$
$x'(1/4) \cdot x'(3/4) = 0$

from which you should be able to find $P_1$ and $P_2$.

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  • $\begingroup$ I think that's what I was looking for. I just have to calculate x'(t), replace t by 1/4 and 3/4 and solve the resulting equations. Thank you very much! $\endgroup$ Commented Nov 16, 2016 at 19:25

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