Find the equations of the tanqent and normal to the parabola

The equation of the given parabola is $y^{2}=4 a x$.

On differentiating $y^{2}=4 a x$ with respect to $x$, we have:

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

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

$\therefore$ The slope of the tangent at $\left(a t^{2}, 2 a t\right)$ is $\left.\frac{d y}{d x}\right]_{\left(a t^{2}, 2 a t\right)}=\frac{2 a}{2 a t}=\frac{1}{t}$.

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

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

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

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

Now, the slope of the normal at $\left(a t^{2}, 2 a t\right)$ is given by,

Thus, the equation of the normal at (at2, 2at) is given as:

$y-2 a t=-t\left(x-a t^{2}\right)$

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

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