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Math Help - Tangent & Normal Q11

  1. #1
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    Tangent & Normal Q11

    The equation of a curve is y=\frac{2x}{3x+1}. The normal to the curve meets the curve at point A(-1,1). Find the equation of the normal.

    I am aware that I have to solve this question using the formula (y-y_1)=\frac{1}{m}(x-x_1).

    However, I am facing problems in simplifying the equation... I would be grateful if someone would show a detailed workings on it, thanks!
    Last edited by Punch; February 17th 2010 at 08:16 AM.
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  2. #2
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    Quote Originally Posted by Punch View Post
    The equation of a curve is y=f(x)=\frac{2x}{3x+1}. The normal to the curve meets the curve at point A(-1,1). Find the equation of the normal.

    I am aware that I have to solve this question using the formula (y-y_1)=\frac{1}{m}(x-x_1).

    However, I am facing problems in simplifying the equation... I would be grateful if someone would show a detailed workings on it, thanks!
    1. Let P\left(p,\ \frac{2p}{3p+1} \right) denote the point on the curve whose normal you want to determine.

    2. The slope of the normal is m = -\frac1{f'(p)}. To calculate f'(x) use the quotient rule:

    f'(x)=\frac2{(3x+1)^2} ..... \implies ..... ..... f'(p)=\frac2{(3p+1)^2} \implies ..... m = -\tfrac12 (3p+1)^2

    3. The slope of the normal is equal to the slope of PA:

    m = \dfrac{\frac{2p}{3p+1} - 1}{p+1}

    4. Both m-values must be equal:

    \dfrac{\frac{2p}{3p+1} - 1}{p+1} = -\tfrac12 (3p+1)^2

    Solve for p. Afterwards determine the equation of the normal.
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