Find the equation of tangent and normal to the parabola $y^2=4ax$ at the point $\left(a{t}^2,2at\right)$

The equation of given curve is,

$y^2=4ax$

Differentiating w.r.t. $x$, we get $2y\frac{dy}{dx}=4a$

$\Rightarrow {\left(\frac{dy}{dx}\right)}_{\left(a{t}^2,2at\right)}=\frac{4a}{4at}=\frac{1}{t}$

Now, the equation of tangent at $\left(a{t}^2,2at\right)$ is

$\frac{y-2at}{x-a{t}^2}={\left(\frac{dy}{dx}\right)}_{\left(a{t}^2,2at\right)}=\frac{1}{t}$

$\Rightarrow (y-2at)t=x-a{t}^2$

$\Rightarrow yt-2a{t}^2=x-a{t}^2\Rightarrow yt=x+a{t}^2$

 hence , it is the required equation of tangent.

and the equation of normal at $\left(a{t}^2,2at\right)$ is,

$\frac{y-2at}{x-a{t}^2}=\frac{-1}{{\left({\displaystyle \frac{dy}{dx}}\right)}_{(a{t}^2,2at)}}$

 $y-2at=-xt+a{t}^3$

hence, $y+xt=2at+a{t}^3$ is the  required equation of normal.