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Math Help - equation of normal to curve

  1. #1
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    equation of normal to curve

    Find the equation of the normal to y = x^2 - 4 which is parallel to x + 3y - 1 = 0

    I found that the gradient of the normal is -1/3 if this is correct, i don't know where to go from here .. if someone could show me the steps please
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  2. #2
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    Re: equation of normal to curve

    Quote Originally Posted by jessm001 View Post
    Find the equation of the normal to y = x^2 - 4 which is parallel to x + 3y - 1 = 0
    I found that the gradient of the normal is -1/3 if this is correct
    If the slope of the normal is \tfrac{-1}{3} then the slope of the tangent at the same point is 3.
    At what point(s) is the slope of the tangent 3~?
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  3. #3
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    Re: equation of normal to curve

    It doesn't say, that's why I got confused, I thought I was missing something
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    Re: equation of normal to curve

    Quote Originally Posted by jessm001 View Post
    Find the equation of the normal to y = x^2 - 4 which is parallel to x + 3y - 1 = 0

    I found that the gradient of the normal is -1/3 if this is correct, i don't know where to go from here .. if someone could show me the steps please
    I think that there is a lack of informations...coordinates of the point that belongs to the curve should be given .
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    Re: equation of normal to curve

    Thanks a lot
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  6. #6
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    Re: equation of normal to curve

    Quote Originally Posted by princeps View Post
    I think that there is a lack of informations...coordinates of the point that belongs to the curve should be given .
    That is not so. There is no lack of information.
    Slope is y'=2x=3. Thus the point is \left( {\tfrac{3}{2},\tfrac{{ - 7}}{4}} \right).
    What is the equation of the normal there?
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