finding the slope of the tangent by differentiating the curve
\(2ay\frac{dy}{dx}=3x^2\)
\(\frac{dy}{dx}=\frac{3x^2}{2ay}\)
m(tangent) at (am2, am3) is \(\frac{3m}{2}\)
normal is perpendicular to tangent so, m1m2 = – 1
m(normal) at (am2, am3) is \(-\frac{2}{3m}\)
equation of normal is given by y – y1 = m(normal)(x – x1)
\(y-am^3=-\frac{2}{3m}(x-am^2)\)