Find the equation of the normal to the curve $a y^{2}=x^{3}$ at the point $\left(a m^{2}, a m^{3}\right)$.
finding the slope of the tangent by differentiating the curve
$2 a y \frac{d y}{d x}=3 x^{2}$
$\frac{d y}{d x}=\frac{3 x^{2}}{2 a y}$
$\mathrm{m}$ (tangent) at $\left(\mathrm{am}^{2}, \mathrm{am}^{3}\right)$ is $\frac{3 \mathrm{~m}}{2}$
normal is perpendicular to tangent so, $m_{1} m_{2}=-1$
$\mathrm{m}$ (normal) at $\left(\mathrm{am}^{2}, \mathrm{am}^{3}\right)$ is $-\frac{2}{3 \mathrm{~m}}$
equation of normal is given by $y-y_{1}=m($ normal $)\left(x-x_{1}\right)$
$y-a m^{3}=-\frac{2}{3 m}\left(x-a m^{2}\right)$
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