Application of Derivatives

Find the slope of the normal at the point (am 3 , am 2 ) to the curve a x 2 = y 3 . - Wired Faculty

Question

Find the slope of the normal at the point (am³ , am² ) to the curve a x² = y³.

Solution

The equation of the curve is a x² = y³Differentiating both sides w.r.t. x, we get,
$2 ax space equals space 3 straight y squared dy over dx space space space space space space space space space space space space space space space space space space space space space space rightwards double arrow space space space space space space dy over dx space equals space fraction numerator 2 ax over denominator space 3 space straight y squared end fraction$
$At space left parenthesis am cubed comma space space space am squared right parenthesis comma space space dy over dx space equals space fraction numerator 2 straight a left parenthesis am cubed right parenthesis over denominator 3 left parenthesis am squared right parenthesis squared end fraction space space equals space fraction numerator 2 straight a squared straight m cubed over denominator 3 straight a squared straight m to the power of 4 end fraction space equals space fraction numerator 2 over denominator 3 space straight m end fraction space therefore space space space space space space space At left parenthesis straight a space straight m cubed comma space space space straight a space straight m squared right parenthesis comma space space space slope space of space tangent space equals space fraction numerator 2 over denominator 3 straight m end fraction therefore space space space space space space At left parenthesis am cubed comma space space am squared right parenthesis comma space space slope space of space normal space space equals space minus fraction numerator 1 over denominator begin display style fraction numerator 2 over denominator 3 straight m end fraction end style end fraction space equals space minus fraction numerator 3 straight m over denominator 2 end fraction$

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