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

We have,
${y^2} = 4ax$ …(i)

Differentiating (i) w.r.t. $x$, we get
$2y\cfrac{{dy}}{{dx}} = 4a \Rightarrow \cfrac{{dy}}{{dx}} = \cfrac{{2a}}{y}$
Therefore, Slope of the tangent at

$(a{t^2},2at)$ is ${\left( {\cfrac{{dy}}{{dx}}} \right)_{(a{t^2},2at)}} = \cfrac{{2a}}{{2at}} = \cfrac{1}{t}$

Hence, the equation of the tangent to (i) at $(a{t^2},\;2at)$ is
$y - 2at = \cfrac{1}{t}(x - a{t^2})$ or $x - ty + a{t^2} = 0$

Slope of normal at $(a{t^2},2at)$ is $\cfrac{{ - 1}}{{{\rm{Slope}}\,\,{\rm{of}}\,{\rm{tangent}}}} = - t$

Therefore the equation of the normal to (i) at $(a{t^2},\;2at)$ is
$y - 2at = - t(x - a{t^2})$ or $tx + y - 2at - a{t^3} = 0$