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Math Help - To find coordinate of the point, need help, please

  1. #1
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    To find coordinate of the point, need help, please

    Find the coordinates of the points on the curve X^2 -2xy+2y^2=4 at which the tangent is parallel to x-axis.
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    Re: To find coordinate of the point, need help, please

    Quote Originally Posted by Minwoo View Post
    Find the coordinates of the points on the curve X^2 -2xy+2y^2=4 at which the tangent is parallel to x-axis.
    Use implicit differentiation to find y^{\prime}.
    Next solve y^{\prime}=0.
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    Re: To find coordinate of the point, need help, please

    Yes, I found dy/dx is (x-y)/(x-2y), then what is the next step? Thank you
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    Re: To find coordinate of the point, need help, please

    Quote Originally Posted by Minwoo View Post
    Yes, I found dy/dx is (x-y)/(x-2y), then what is the next step? Thank you
    At what points on the curve is it true that \frac{dy}{dx}=0~?
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  5. #5
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    Re: To find coordinate of the point, need help, please

    Sorry, I really dunno
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    Re: To find coordinate of the point, need help, please

    Quote Originally Posted by Minwoo View Post
    Sorry, I really dunno
    Are you saying that you cannot solve \frac{x-y}{x-2y}=0~?

    If you are, then you have no business even trying this problem!
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  7. #7
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    Re: To find coordinate of the point, need help, please

    You don't really have to find the derivative explicitely. Going back to your original equation, x^2 -2xy+2y^2=4, implicit differentiation gives 2x- 2y- 2xy'z+ 4yy'= 0. Now, because you are looking for points where y'= 0, put y'= 0 into that equation: 2x- 2y= 0. That tells you that y= x. Replace y with x in the original equation, x^2 -2xy+2y^2=4 to get an equation in x only and solve that for x.
    Last edited by HallsofIvy; September 7th 2012 at 06:31 AM.
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  8. #8
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    Re: To find coordinate of the point, need help, please

    Thank you for the help
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