Given as ay2 = x3
Differentiate with respect to x, to get slope of tangent 2ay(dy/dx) = 3x2
dy/dx = 3x2/2ay
m(tangent) at (am2, am3) is 3m/2
The normal is perpendicular to tangent therefore, m1m2 = – 1
m(normal) at (am2, am3) is -2/3m
The equation of normal is given by y – y1 = m(normal)(x – x1)
y - am3 = (-2/3m)(x - am2)