# Find the equation of the normal at the point $left(a m^{2}, a m^{3}ight)$ for the curve $a y^{2}=x^{3}$.

Question:

Find the equation of the normal at the point

Solution:

The equation of the given curve is $a y^{2}=x^{3}$.

On differentiating with respect to x, we have:

$2 a y \frac{d y}{d x}=3 x^{2}$

$\Rightarrow \frac{d y}{d x}=\frac{3 x^{2}}{2 a y}$

The slope of a tangent to the curve at $\left(x_{0}, y_{0}\right)$ is $\left.\frac{d y}{d x}\right]_{\left(x_{0}, y_{0}\right)}$.

$\Rightarrow$ The slope of the tangent to the given curve at $\left(a m^{2}, a m^{3}\right)$ is

$\left.\frac{d y}{d x}\right]_{\left(a m^{2}, a m^{3}\right)}=\frac{3\left(a m^{2}\right)^{2}}{2 a\left(a m^{3}\right)}=\frac{3 a^{2} m^{4}}{2 a^{2} m^{3}}=\frac{3 m}{2} .$

$\therefore$ Slope of normal at $\left(a m^{2}, a m^{3}\right)=\frac{-1}{\text { slope of the tangent at }\left(a m^{2}, a m^{3}\right)}=\frac{-2}{3 m}$

Hence, the equation of the normal at $\left(a m^{2}, a m^{3}\right)$ is given by,

$y-a m^{3}=\frac{-2}{3 m}\left(x-a m^{2}\right)$

$\Rightarrow 3 m y-3 a m^{4}=-2 x+2 a m^{2}$

$\Rightarrow 2 x+3 m y-a m^{2}\left(2+3 m^{2}\right)=0$