Chiming in with Fred, I didn't understand the question, either. The general planar conic section is expressible in the implicit form: A x^2 + B xy + C y^2 + Dx + Ey + F = 0 (where ABC <> 0). Maybe you could phrase your question by using mathematical notation so that it's crystal clear just what you're asking. --Dan P.S. (For t in R u {oo}) the parametric equations (x(t),y(t)) = ((t^2 - 1)/(t^2 + 1), 2t/(t^2 + 1)) yield a circle in the plane. But (x(t),y(t),z(t)) = (t^2 - 1, 2t, t^2 + 1)) define the space curve that's the intersection of the cone x^2 + y^2 = z^2 with the parabolic cylinder x = (1/4)y^2 - 1, with all coordinates unbounded. I'm not sure what that's called. << AIUI, in order to form conic setion curves using rational parametric polynomials such that the (Cartesian) coordinates of the loci, the radii and the coefficients are all in Q, the polymonials have to be at least quintic. Otherwise, at least one of the loci, radii or coefficients are radicals. Presuming that is true, is there any relation between that and the fact that quintics are the lowest degree polys which are not always solvable in terms of radicals (given coefficients in Q)?

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Sometimes the brain has a mind of its own.