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Math Help - Locus of point

  1. #1
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    Locus of point

    The tangent at the point (t,1/t) on the hyperbola xy=1 meets the x and y axes at A and B respectively. The point C on the line AB is such that AC:AB=a:b. prove that the locus of C as t varies is the rectangular hyperbola
    xy=\frac{4ab}{(a+b)^2}

    I know A(2t,0) and B(0,2/t). So i let C(x,y) and found the values of AC and BC, then used the ratio \frac{a}{b}=\frac{AC}{BC}
    I got
    \frac{a^2}{b^2}=\frac{x^2-4tx+4t^2+y^2}{x^2+y^2-\frac{4y}{t}+\frac{4}{t^2}}
    But now i don't know what to do next, Don't see any ways to make the given equation.
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  2. #2
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    Quote Originally Posted by arze View Post
    The tangent at the point (t,1/t) on the hyperbola xy=1 meets the x and y axes at A and B respectively. The point C on the line AB is such that AC:AB=a:b. prove that the locus of C as t varies is the rectangular hyperbola
    xy=\frac{4ab}{(a+b)^2}

    I know A(2t,0) and B(0,2/t). So i let C(x,y) and found the values of AC and BC, then used the ratio \frac{a}{b}=\frac{AC}{BC}
    I got
    \frac{a^2}{b^2}=\frac{x^2-4tx+4t^2+y^2}{x^2+y^2-\frac{4y}{t}+\frac{4}{t^2}}
    But now i don't know what to do next, Don't see any ways to make the given equation.
    1. You didn't use the fact that C must be placed on the tangent line with the equation:

    y = -\frac1t^2 x + \frac2t

    2. Sub in the term for y into the equation:

    \frac{a^2}{b^2}=\frac{(x-2t)^2+y^2}{x^2+\left(y-\frac2t\right)^2}

    and solve for x (I've got x=\frac{2bt}{a+b}~\vee~x=\frac{2bt}{b-a})

    3. Solve this equation for t and plug in the result into the equation of the tangent line. After expanding the brackets you'll get the given result.
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  3. #3
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    I can't seem to get the values for x.
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  4. #4
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    ok! got it now! thanks
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