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Thread: Find the equation of tangent and normal to the curve

  1. #1
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    Find the equation of tangent and normal to the curve

    Question : Find the eq of tangent and normal to the curve
    $\displaystyle y(x-2)(x-3)-x+7 = 0 $
    at the point where the curve meets the x-axis
    -----------------------------------------------------------------
    Solution:

    $\displaystyle y = \frac{x-7}{x^2 - 5x +6 }$.............Is that correct?

    $\displaystyle \therefore \frac{dy}{dx} = \frac{-x^2 + 14x -29}{(x^2 - 5x +6)^2}$...................Is that correct?????


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  2. #2
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    Quote Originally Posted by zorro View Post
    Question : Find the eq of tangent and normal to the curve
    $\displaystyle y(x-2)(x-3)-x+7 = 0 $
    at the point where the curve meets the x-axis
    -----------------------------------------------------------------
    Solution:

    $\displaystyle y = \frac{x-7}{x^2 - 5x +6 }$.............Is that correct?

    $\displaystyle \therefore \frac{dy}{dx} = \frac{-x^2 + 14x -29}{(x^2 - 5x +6)^2}$...................Is that correct?????

    differentiate (x - 7)/(x^2 - 5x + 6) - Wolfram|Alpha

    You can do the simplifying of the Wolfram expression and see if it's equivalent to what you got.
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  3. #3
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    what should i do nxt

    Quote Originally Posted by mr fantastic View Post
    differentiate (x - 7)/(x^2 - 5x + 6) - Wolfram|Alpha

    You can do the simplifying of the Wolfram expression and see if it's equivalent to what you got.

    $\displaystyle \frac{dy}{dx} = \frac{- (x^2 - 14x + 31)}{(x^2 - 5x + 6)^2}$
    what should i do??????
    Now eg of tangent is at $\displaystyle (x_0,y_0)$ is given by

    $\displaystyle (y-y_0) = \frac{dy}{dx} (x-x_0)$

    but i dont know $\displaystyle (x_0,y_0)$.................what should i do??????
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  4. #4
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    Quote Originally Posted by zorro View Post
    $\displaystyle \frac{dy}{dx} = \frac{- (x^2 - 14x + 31)}{(x^2 - 5x + 6)^2}$
    what should i do??????
    Now eg of tangent is at $\displaystyle (x_0,y_0)$ is given by

    $\displaystyle (y-y_0) = \frac{dy}{dx} (x-x_0)$

    but i dont know $\displaystyle (x_0,y_0)$.................what should i do??????
    When the curve meets the x-axis, y = 0. Therefore $\displaystyle y_0 = 0$. To get $\displaystyle x_0$, solve $\displaystyle -x + 7 = 0$ (I hope you see why).
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