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Keavon
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| let a = self.p0.to_vec2() - 2.0 * self.p1.to_vec2() + self.p2.to_vec2(); | ||
| let b = 2.0 * (self.p1.to_vec2() - self.p0.to_vec2()); | ||
| let c = self.p0.to_vec2() - p.to_vec2(); | ||
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| // coefficients of x(t) \cross x'(t) | ||
| let c2 = a.cross(b); | ||
| let c1 = -2.0 * c.cross(a); | ||
| let c0 = b.cross(c); |
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I'd like a double check on the math here, both in terms of its correctness (I'm not certain of it) and its formatting (is it more faithful to the canonical Bézier equations to reverse the orders of the terms, swap the order of the cross product and negate it, etc.?). This mostly arose from pattern recognition based fiddling and I wasn't able to quite grasp its relationship with the Bézier equations on Wikipedia. So please confirm this is right.
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Follows up #288 which added
tangents_to_point()toCubicBez, by adding it toQuadBeznow as well.