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Math Help - Parametric equations of a general equation of parabola

  1. #1
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    Parametric equations of a general equation of parabola

    A parabola passes thorough the points (1,0) & (0,1) and x & y axes are tangent at these points respectively. How to find the parametric equations to such a parabola?

    I have derived the equation to the parabola, which is of the "general equation of a parabola" type Ax^2+Bxy+Cy^2+Dx+Ey+F=0, not y^2=4ax type.

    The general parametric equations of a parabola are
    x=at^2+bt+c
    y=lt^2+mt+n

    How to go about it?
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  2. #2
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    Well, what you have Ax^2+ Bxy+ Cy^2+ Dx+ Ey+ F= 0 is NOT the "general equation of a parabola", it is the general equation of a "conic section" and so includes hyperbolas, ellipses, and circles (as well as "degenerate conic sections", straight lines, pairs of straight lines, and single points) so you can't go from that equation to parametric equations for a parabola.
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  3. #3
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    I said its of the "type", not that it is exactly so.
    Let me write it completely (thought it to be understood)
    Ax^2+Bxy+Cy^2+Dx+Ey+F=0
    Provided B^2=4AC

    So I cannot derive parametric equations for such a parabola. Is that the verdict?
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  4. #4
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    Done. Solved it.

    Equation of the parabola x^2+y^2-2xy-2x-2y+1=0

    Parametric equations:
    y=\left(\frac {t+1} {2}\right)^2
    x=\left(\frac {t-1} {2}\right)^2

    If anyone needs the steps let me know.
    Last edited by ontherocks; April 20th 2010 at 10:19 PM.
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