Find the equation

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

$\frac{\mathrm{dx}}{\mathrm{dt}}=2 \mathrm{at}$

$\frac{\mathrm{dy}}{\mathrm{dt}}=2 \mathrm{a}$

Now dividing $\frac{\mathrm{dy}}{\mathrm{dt}}$ and $\frac{\mathrm{dx}}{\mathrm{dt}}$ to obtain the slope of tangent

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{\mathrm{t}}$

$\mathrm{m}$ (tangent) at $\mathrm{t}=1$ is 1

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

$\mathrm{m}($ normal $)$ at $\mathrm{t}=1$ is $-1$

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

$y-2 a=1(x-a)$

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

$y-2 a=-1(x-a)$