Find the equation of the normal at the point `(a m^2,a m^3)`for the curve `a y^2=x^3`.

Explanation

Equation of curve
` ay^(2) = x^(3)`
` rArr 2ay (dy)/(dx) = 3x^(2)`
` rArr (dy)/(dx) = (3x^(2))/(2ay)`
Slope of tangent at point `(am^(2), am^(3)) = (3(am^(2)^(2))/(2a(am^(3))) = (3m)/2`
` rArr" Slope of normal "=(-2)/(3m)`
and equation of normal
` y-am^(3) = - 2/(3m) (x- am^(2))`
` rArr 3my - 3am^(4) = - 2x + 2am^(2)`
` rArr 2x + 3my - 3am^(4) - 2am^(2) = 0`