Find the equation

finding the slope of the tangent by differentiating the curve

$2 y \frac{d y}{d x}=4 a$

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{2 \mathrm{a}}{\mathrm{y}}$

$\mathrm{m}$ (tangent) at $\left(\frac{\mathrm{a}}{\mathrm{m}^{2}}, \frac{2 \mathrm{a}}{\mathrm{m}}\right)$

$\mathrm{m}$ (tangent) $=\mathrm{m}$

normal is perpendicular to tangent so, $m_{1} m_{2}=-1$

$\mathrm{m}$ (normal) $=-\frac{1}{\mathrm{~m}}$

equation of tangent is given by $y-y_{1}=m($ tangent $)\left(x-x_{1}\right)$

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

equation of normal is given by $y-y_{1}=m($ normal $)\left(x-x_{1}\right)$

$y-\frac{2 a}{m}=-\frac{1}{m}\left(x-\frac{a}{m^{2}}\right)$