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Math Help - intersection of a line and a curve

  1. #1
    Member helloying's Avatar
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    intersection of a line and a curve

    Find the coordinates of the points of itersection A and B of the line 2x+y+2=0 and the curve 1/x +2/y= 1/2

    i reform the eqn of the line to be y=-2-2x and the curve is y= \frac{2}{(0.5-1/x)}

    then i sub the two eqn and i 2=(0.5-1/x)(-2-2x)

    and so x=-2. but i only got one ans and the qn say there are two point of intersection. How to solve this qn?
    Last edited by helloying; July 4th 2009 at 09:33 PM.
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  2. #2
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    Quote Originally Posted by helloying View Post
    Find the coordinates of the points of itersection A and B of the line 2x+y+2=0 and the curve 1/x +2/y= 1/2

    i reform the eqn of the line to be y=-2-2x and the curve is \frac{2}{(0.5-1/x)}

    then i sub the two eqn and i 2=(0.5-1/x)(-2-2x)

    and so x=-2. but i only got one ans and the qn say there are two point of intersection. How to solve this qn?
    y=-2-2x, and substitute to get \frac{1}{x}+\frac{2}{-2-2x}=\frac{1}{2}
    \frac{1}{x}-\frac{1}{1+x}=\frac{1}{2}

    \frac{1+x}{x(1+x)}-\frac{x}{x(1+x)}=\frac{1}{2}

    \frac{1}{x+x^2}=\frac{1}{2}

    x+x^2=2
    x^2+x-2=(x+2)(x-1)=0

    x=1 and x=-2 and therefore y=-4 and y=2

    so (1,-4) and (-2,2)
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  3. #3
    Super Member malaygoel's Avatar
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    Quote Originally Posted by helloying View Post
    2=(0.5-1/x)(-2-2x)
    opening brackets,
    2=-1-x+\frac{2}{x}+2

    2x=-x-x^2+2+2x

    which gives
    x=-2
    and x=1
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