Application of Derivatives

Find the equation of the normal at the point (am 2 , am 3 ) for the curve ay 2 = x 3 . - Wired Faculty

Question

Find the equation of the normal at the point (am², am³) for the curve ay²= x³.

Solution

The equation of the curve is a y² = x³Differentiating both sides w.r.t. x,
$2 ay dy over dx space equals space 3 straight x squared space space space space space space space space space space space space space space or space space space space space dy over dx space equals space fraction numerator 3 straight x squared over denominator 2 ay end fraction At space left parenthesis am squared comma space space am cubed right parenthesis comma space space space dy over dx space equals space fraction numerator 3 space cross times space straight a squared space straight m to the power of 4 over denominator 2 space straight a space cross times space straight a space straight m cubed end fraction space equals space fraction numerator 3 straight a squared straight m to the power of 4 over denominator 2 straight a squared straight m cubed end fraction space equals space 3 over 2 straight m which space is space slope space of space tangent. therefore space space space space space space space space slope space of space normal space space equals space fraction numerator 2 over denominator 3 straight m end fraction therefore space space space space space space the space equation space of space normal space at space left parenthesis am to the power of minus comma space space space space am cubed right parenthesis space is space space space space space space space space space space space space space space space space space space straight y minus straight a space straight m cubed space equals space minus fraction numerator 2 over denominator 3 straight m end fraction left parenthesis straight x minus am cubed right parenthesis space space space or space space space space space space 3 my minus 3 am to the power of 4 space equals space minus 2 straight x plus 2 am squared or space space space space space 2 straight x plus 3 my minus 3 am to the power of 4 minus 2 am squared space equals space 0 or space space space space space space 2 straight x plus 3 my minus am squared left parenthesis 3 straight m squared plus 2 right parenthesis space equals space 0$

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