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Math Help - rectangular hyperbola

  1. #1
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    Cool rectangular hyperbola

    Hi folks,

    I am struggeling with the following problem:

    The tangent at the point P(ct, c/t) (t>0) on the hyperbola xy = c^2 meets the x-axis at A and the y-axis at B. The normal at P to the rectangular hyperbola meets the line y = x at C and the line y = -x at D.

    The normal at P meets the hyperbola again at Q and the mid point of PQ is M. Prove that as t varies, the point M lies on the curve c^2(x^2 - y^2)^2 + 4x^3y^3 = 0

    I have calculated the points as follows: A(2ct,0), B(0, 2c/t), C(\frac{c(t^2 + 1)}{t}, \frac{c(t^2 + 1)}{t}), D(\frac{c(t^2 - 1)}{t}, \frac{-c(t^2 - 1)}{t})

    I have the point Q at (\frac {-c}{t^3}, -ct^3) and the mid point of PQ, point M at (\frac{c(t^4 - 1)}{2t^3}, \frac{-c(t^4 - 1)}{2t})

    The locus of M is usually found by eliminating T from the coordinates of the point, but I can't seem to get anywhere. I end up with t^2 = \frac{-y}{x} and we have xy = c^2 but I can't eliminate t completely. Should I try working back from the given result? I feel there has to be a better way!
    Last edited by s_ingram; August 7th 2009 at 06:19 AM. Reason: Latex syntax error
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  2. #2
    Super Member malaygoel's Avatar
    Joined
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    you have
    x=\frac{c(t^4 - 1)}{2t^3}
    y=\frac{-c(t^4 - 1)}{2t}
    hence,
    \frac{y}{x}=-t^2....(1)
    xy=\frac{c^2(t^4-1)^2}{4t^4}....(2)

    Sustituting the value of t^2 from eq(1) to eq(2)
    xy=\frac{c^2\{(\frac{-y}{x})^2-1\}^2}{4(\frac{-y}{x})^2}
    Last edited by malaygoel; August 7th 2009 at 08:06 PM.
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