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    A curve is defined by the parametric equation; x= t^3 / (t^2 +1) , y= t^2 / (t^2 +1) .

    Show that the cartesian equation for the curve is y^3=x^2(1-y).

    If someone could show me how to get this Cartesian equation, I will really appreciate it. Thanks
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    by inspection you can see that

    x = yt

    so

    y = x/t

    y3 = {x/t}3

    y3 = {x3}
    /{t3}

    y3 = {x2}*x/ /{t3}

    see if you can continue...
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    (Original post by the bear)
    by inspection you can see that

    x = yt

    so

    y = x/t

    y3 = {x/t}3

    y3 = {x3}
    /{t3}

    y3 = {x2}*x/ /{t3}

    see if you can continue...

    Thank you, I have tried this but I cannot get x/t^3 into (1-y). Could you please show me?
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    (Original post by AjayTaak)
    Thank you, I have tried this but I cannot get x/t^3 into (1-y). Could you please show me?
    i would start things off by rewriting y:

    \displaystyle y= \frac{t^2}{t^2 +1} = \frac{t^2+1-1}{t^2+1} = 1 - \frac{1}{t^2+1}

    (You could do this using other division methods)

    That should help you see where 1-y comes from in the solution.
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    (Original post by AjayTaak)
    Thank you, I have tried this but I cannot get x/t^3 into (1-y). Could you please show me?
    x/t3 can be written as...

    { t3/{ 1 + t2 } }/t3...

    ====> 1/{t2 + 1 }

    we need to show that

    1 - y = 1/{t2 + 1 }

    if you put in t2/{ 1 + t2 } for y the result emerges as required...
 
 
 
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