20) Find the equation of the normal at the point ( am^2 , am^3 ) for the curve ay ^2 = x ^3.

Answers (1)

Given equation of curve is 
ay ^2 = x ^3\Rightarrow y^2 = \frac{x^3}{a}
Slope of tangent

 2y\frac{dy}{dx} = \frac{3x^2 }{a} \Rightarrow \frac{dy}{dx} = \frac{3x^2}{2ya}
at point ( am^2 , am^3 )
\frac{dy}{dx} = \frac{3(am^2)^2}{2(am^3)a} = \frac{3a^2m^4}{2a^2m^3} = \frac{3m}{2}
Now, we know that 
Slope \ of \ normal = \frac{-1}{Slope \ of \ tangent} = \frac{-2}{3m}
equation of normal at point ( am^2 , am^3 ) and with slope \frac{-2}{3m}
y-y_1=m(x-x_1)\\ y-am^3 = \frac{-2}{3m}(x-am^2)\\ 3ym - 3am^4 = -2(x-am^2)\\ 3ym +2x= 3am^4+2am^2
Hence, the equation of normal is 3ym +2x= 3am^4+2am^2

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