TITLE: Quadratic Bezier curves representation as implicit quadratic equation
QUESTION [5 upvotes]: A quadratic bezier curve from points P1=(x1, y1) to P3=(x3, y3) with control point P2=(x2, y2) can be represented as parametric quadratic curve P(t) where t is in [0, 1].
$$P(t) = (P_1t + P_2(1 - t))t + (P_2t + P_3(1 - t))(1 - t)
         = P_1t^2 + P_2(1 - t)t + P_3(1 - t)^2$$
We can extend the range of t to be all real numbers, from minus infinity to plus infinity, and then we will get a nice curve dividing the plane in two.
I read somewhere, that this infinete curve is either a straight line (if P1, P2 and P3 are colinear) or a parabola. Could you please confirm that?
Since division of plane by a parabola or a line can be expressed by quadratic equation of form:
Point P=(x, y) is 
  on one side  if F(P) > 0
  on the other if F(P) < 0
  on the curve if F(P) == 0
where F(P) = A * x^2 + B * y^2 + C * x * y + D * x + E * y + F

How can we, starting from points P1, P2, P3 calculate the numbers A, B, C,D, E, F?
I think something similar is done in this wolfram demostration http://demonstrations.wolfram.com/FocusAndDirectrixInAQuadraticBezierCurve/
Thank you

REPLY [3 votes]: The point $(x,y)$ is on the quadratic Bezier curve if the equations
$$
  x1*t^2 + x2*t*(1-t)*2 + x3*(1-t)^2 = x \\
  y1*t^2 + y2*t*(1-t)*2 + y3*(1-t)^2 = y
$$
have a common solution, $t$. There is an extensive theory (mostly lost in the mists of time) that tells us when two polynomials have a common root. Look for references on elimination theory and resultants. Specific resultants bear the names of Dixon, Sylvester, and others. A fairly recent reference is here. Resultants are attractive because they show some structure -- the formulae look much nicer than the mass of symbols that you get by brute-force elimination using programs like Alpha or Sage.