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    Hi,

    The question is as follows:

    Let P be a point on the curve x=t², y=1/t. If the tangent to the curve at P meets the x-axis and y-axis at A and B respectively, prove that PA=2BP.
    Seems simple enough - but I end up with PA² = 4t4 + 1/t² and BP² = t4 + 9/4t², the roots of which do not satisfy the requirement.

    Could someone show me how it ought to be done?

    Cheers!
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    Let P be a point on the curve x=t², y=1/t. If the tangent to the curve at P meets the x-axis and y-axis at A and B respectively, prove that PA=2BP.
    \frac{dx}{dt}=2t, \frac{dy}{dt}=-\frac{1}{t^2} \rightarrow \frac{dy}{dx}=\frac{dy}{dt}.\fra  c{dt}{dx}=-\frac{1}{t^2}.\frac{1}{2t}=-\frac{1}{2t^3}
    Let P be the point (p^2,\frac{1}{p}) where t=p.
    Then at P,x=p^2, y=\frac{1}{p}, \frac{dy}{dx}=-\frac{1}{2p^3}
    Thus the tangent at P has equation: y-\frac{1}{p}=-\frac{1}{2p^3}(x-p^2)
    Or simplified:2p^3 y + x = 3p^2
    At A, when the tangent meets the x axis (y=0) then x=3p^2
    At B, when the tangent meets the y axis (x=0) then y=\frac{3}{2p}
    Thus we have the information:
    A=(3p^2,0)
    B=(0,\frac{3}{2p})
    P=(p^2, \frac{1}{p})
    If we switch to a "vector style" approach:
    PA=OA-OP=i(3p^2-p^2)+j(0-\frac{1}{p})+0k=i(2p^2)+j(-\frac{1}{p})+0k
    BP=OP-OB=i(p^2-0)+j(\frac{1}{p}-\frac{3}{2p})+0k=i(p^2)+j(-\frac{1}{2p})+0k
    Then PA=i(2p^2)+j(-\frac{1}{p})+0k and 2BP=i(2p^2)+j(-\frac{1}{p})+0k
    That is, PA=2BP.

    Note I've treated PA and BP as vectors because there isn't a modulus sign. In any case, the above still shows |PA|=2|BP| as the vectors PA and 2BP are exactly the same (thus having the same length).

    If you haven't looked at vectors yet then you can think of PA=i(2p^2)+j(-\frac{1}{p})+0k as increasing the x coordinate by 2p^2 and decreasing the y coordinate by \frac{1}{p} as you move from P to A. Note that since we're working in 2 dimensions the coefficient of k doesn't change.
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    (Original post by Theophilus7)
    Hi,

    The question is as follows:



    Seems simple enough - but I end up with PA² = 4t4 + 1/t² and BP² = t4 + 9/4t², the roots of which do not satisfy the requirement.

    Could someone show me how it ought to be done?

    Cheers!
    im bored so will do it using co-ord geom.
    using the co-ords from post 2
    A=(3p^2,0)
    B=(0,3/2p)
    general point on curve has co-ords (p^2,1/p)
    PA^2=4p^4+1/p^2=(4p^6+1)/p^2
    BP^2=p^4+(1/4p^2)=(4p^6+1)/4p^2
    PA=rt(4p^6+1)/p
    BP=rt(4p^6+1)/2p
    giving PA=2BP
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    Thank you both . :-)
 
 
 
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Updated: August 27, 2005

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