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? asked in Science & MathematicsMathematics · 10 years ago

Rational parametrization of A x^2 + B xy + C y^2 + D x + E y + F = 0 with A, B, C not all zero?

I'm also very interested in the derivation.

Update:

Hi Scythian,

Yes, a rational parametrization doesn't give every point, but give some points precisely.

For example (non-unique),

x^2 + y^2 = 1.................... [ 2t/(1 + t^2) , (1 - t^2)/(1 + t^2) ].

x^2 / a^2 + y^2 / b^2 = 1.....[ a 2t/(1 + t^2) , b (1 - t^2)/(1 + t^2) ]

x^2 / a^2 - y^2 / b^2 = 1......[ a (1 + t^2)/2t , b (1 - t^2)/2t ]

But "any generalized parametric function" would be great.

Rational Curves http://www.cs.mtu.edu/~shene/COURSES/cs3621/NOTES/...

1 Answer

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  • 10 years ago
    Favorite Answer

    If you mean converting f(x,y) = 0 into the parametric equations (x(t),y(t)), where x(t), y(t) are rational functions of t (i.e., yield rational values for rational t), then that implies that for any point on the curve (x,y), if x is rational, so is y. This is generally not true. For example, there is no rational parametrization of the circle. We can have (Sin(t), Cos(t)) as a parametric function of the circle, but it's not a rational one.

    Maybe explain better what's meant by "rational parametrization"? How about any generalized parametric function that works for all A x^2 + B xy + C y^2 + D x + E y + F = 0?

    Edit: Thanks for the reference, Jin, I'll look into this to get a clearer idea. Leave this one open.

    Edit 2: Okay, I understand better now what is meant by "rational parametrization", and this is an interesting problem, but don't expect an answer anytime soon. But one of the forms of parametrization happens to be the same I employed in answering this question, in coming up with the "generalized parameter a":

    http://answers.yahoo.com/question/index;_ylt=AtybR...

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