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    Hi, I've been doing some work on parametric equations and I have a few questions which I'm stuck on. I would appreciate any help given!

    With Q3b, is there are way of completing the question without finding the cartesian equation (or is it a necessary step?). And I'm not too sure what I should do next, as I can't factorise it.

    For Q4c, I think my working out so far is correct (parts a/b are correct) but I don't see how I can prove PA=2BP as they're in different forms.



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    (Original post by Bazinga?)
    With Q3b, is there are way of completing the question without finding the cartesian equation (or is it a necessary step?). And I'm not too sure what I should do next, as I can't factorise it.
    Yes. In the equation of the tangent, (is it x+3y-4=0 ?) replace x with t^3 and y with t^2 to get a cubic in t. You know that t=-2 is a repeated root so finding the other root should be easy.
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    (Original post by BabyMaths)
    Yes. In the equation of the tangent, (is it x+3y-4=0 ?) replace x with t^3 and y with t^2 to get a cubic in t. You know that t=-2 is a repeated root so finding the other root should be easy.
    Oh okay, by doing what you said, I got the right answer. However, I don't understand why you substitute the t's back into the equation of the tangent? And how do you know t=-2 is a repeated root?

    Also, I know it's more long winded, but if I found the cartesian equation of the curve and equated that to the tangent, would that find t?
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    (Original post by Bazinga?)
    Hi, I've been doing some work on parametric equations and I have a few questions which I'm stuck on. I would appreciate any help given!

    With Q3b, is there are way of completing the question without finding the cartesian equation (or is it a necessary step?). And I'm not too sure what I should do next, as I can't factorise it.

    For Q4c, I think my working out so far is correct (parts a/b are correct) but I don't see how I can prove PA=2BP as they're in different forms.



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    For 3)
    Your Cartesian
    \displaystyle \frac{1}{3}x+x^{\frac{2}{3}}-\frac{4}{3}=0
    sub \displaystyle x^3=t and multiply by 3
    \displaystyle t^3+3t^2-4=0
    \displaystyle t^3-t^2+4t^2-4=0
    factorizing by pair
    \displaystyle t^2(t-1)+4(t^2-1)=0
    \displaystyle t^2(t-1)+4(t-1)(t+1)=0
    factorizing
    \displaystyle (t-1)(t^2+4t+4)=0
    \displaystyle (t-1)(t+2)^2=0
    and here is the repeated root and the other one.

    An other method maybe the long division by (t+2) because you know
    that one of the roots is -2 (where the tangent meets the curve)
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    (Original post by ztibor)
    For 3)
    Your Cartesian
    \displaystyle \frac{1}{3}x+x^{\frac{2}{3}}-\frac{4}{3}=0
    sub \displaystyle x^3=t and multiply by 3
    \displaystyle t^3+3t^2-4=0
    \displaystyle t^3-t^2+4t^2-4=0
    factorizing by pair
    \displaystyle t^2(t-1)+4(t^2-1)=0
    \displaystyle t^2(t-1)+4(t-1)(t+1)=0
    factorizing
    \displaystyle (t-1)(t^2+4t+4)=0
    \displaystyle (t-1)(t+2)^2=0
    and here is the repeated root and the other one.

    An other method maybe the long division by (t+2) because you know
    that one of the roots is -2 (where the tangent meets the curve)
    Okay, thank you for your help Would you be able to help with 4?
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    (Original post by Bazinga?)
    Okay, thank you for your help Would you be able to help with 4?
    Prove that
    \frac{PA}{PB}=2

    for c)
    \displaystyle BP=\sqrt{p^4+\frac{1}{4p^2}}
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    (Original post by ztibor)
    Prove that
    \frac{PA}{PB}=2

    for c)
    \displaystyle BP=\sqrt{p^4+\frac{1}{4p^2}}
    Thanks for your help, it is very much appreciated!
 
 
 
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