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Thread: Tangent & Normal Q11

  1. #1
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    Tangent & Normal Q11

    The equation of a curve is $\displaystyle y=\frac{2x}{3x+1}$. 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 $\displaystyle (y-y_1)=\frac{1}{m}(x-x_1)$.

    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; Feb 17th 2010 at 07:16 AM.
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  2. #2
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    Quote Originally Posted by Punch View Post
    The equation of a curve is $\displaystyle y=f(x)=\frac{2x}{3x+1}$. 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 $\displaystyle (y-y_1)=\frac{1}{m}(x-x_1)$.

    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 $\displaystyle P\left(p,\ \frac{2p}{3p+1} \right)$ denote the point on the curve whose normal you want to determine.

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

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

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

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

    4. Both m-values must be equal:

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

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