Find the equation

finding slope of the tangent by differentiating $x$ and $y$ with respect to $t$

$\frac{\mathrm{dx}}{\mathrm{dt}}=\frac{\left(1+\mathrm{t}^{2}\right) 4 \mathrm{at}-2 \mathrm{at}^{2}(2 \mathrm{t})}{\left(1+\mathrm{t}^{2}\right)^{2}}$

$\frac{\mathrm{dx}}{\mathrm{dt}}=\frac{4 \mathrm{at}}{\left(1+\mathrm{t}^{2}\right)^{2}}$

$\frac{d y}{d t}=\frac{\left(1+t^{2}\right) 6 a t^{2}-2 a t^{3}(2 t)}{\left(1+t^{2}\right)^{2}}$

$\frac{d y}{d t}=\frac{6 a t^{2}+2 a t^{4}}{\left(1+t^{2}\right)^{2}}$

Now dividing $\frac{d y}{d t}$ and $\frac{d x}{d t}$ to obtain the slope of tangent

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{6 \mathrm{at}^{2}+2 \mathrm{at}^{4}}{4 \mathrm{at}}$

$\mathrm{m}$ (tangent) at $\mathrm{t}=\frac{1}{2}$ is $\frac{13}{16}$

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

$\mathrm{m}($ normal $)$ at $\mathrm{t}=\frac{1}{2}$ is $-\frac{16}{13}$

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

$y-\frac{a}{5}=\frac{13}{16}\left(x-\frac{2 a}{5}\right)$

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

$y-\frac{a}{5}=-\frac{16}{13}\left(x-\frac{2 a}{5}\right)$