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Math Help - Equation of line normal to curve

  1. #1
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    Equation of line normal to curve

    Hi. Equation for curve:
    y = 6x - 2x^2 at (2,4)

    My work:
    = 6 - 4x
    = 6 - 4(2)
    = 6 - 8 = -2 = M (slope)

    reciprocal: 1/-2
    y - y1 = m (x - x1)
    y - 4 = -2(x-2)
    y - 4 = -2x + 4
    y = -2x

    but this is wrong? thanks.
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  2. #2
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    Re: Equation of line normal to curve

    The normal line is the line perpindicular to the curve, and that requires the negative reciprocal I think. So the slope of the normal line should be just 1/2. You can type the question into WolframAlpha to confirm.
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  3. #3
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    Re: Equation of line normal to curve

    got it. thanks
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  4. #4
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    Re: Equation of line normal to curve

    Quote Originally Posted by togo View Post
    Hi. Equation for curve:
    y = 6x - 2x^2 at (2,4)

    My work:
    = 6 - 4x
    = 6 - 4(2)
    = 6 - 8 = -2 = M (slope)

    reciprocal: 1/-2
    y - y1 = m (x - x1)
    y - 4 = -2(x-2)
    y - 4 = -2x + 4
    y = -2x

    but this is wrong? thanks.
    It's already been established that the gradient of your normal is 1/2, not -1/2. But I'm wondering why you substituted -2 as m?
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