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Application Of Derivatives

Question
CBSEENMA12035124

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

Solution
The equation of the curve is a y2 = x3
Differentiating 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 or space space space space space space dy over dx space equals space fraction numerator 3 straight x squared over denominator 2 ay end fraction
At left parenthesis am squared comma space space am cubed right parenthesis comma space space dy over dx space equals space fraction numerator 3 cross times space straight a squared straight m to the power of 4 over denominator 2 straight a cross times space am 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 is slope of tangent.
therefore 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 the space equation space of space normal space at space left parenthesis am to the power of minus comma space space space am cubed right parenthesis space is
space space space space space space space space space space straight y minus am 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 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