Find the equation of tangent and normal to the parabola y² = 6x at the positive and of latus rectum.

Given parabola y² = 6x
$4\mathrm{a}=6\Rightarrow \mathrm{a}=\frac{3}{2}$
Positive end of the latusrectum $L=(\mathrm{a},2\mathrm{a})=\left(\frac{3}{2},3\right)$
Equation of tangent to the parabola ${\mathrm{y}}^2=6\mathrm{x} \mathrm{at} \mathrm{L}\text{ is }{\mathrm{S}}_1=0$
$\begin{array}{l}{\mathrm{yy}}_1=3\left(\mathrm{x}+{\mathrm{x}}_1\right)\text{ at point }\left(\frac{3}{2},3\right)\\ 3\mathrm{y}=3\left(\mathrm{x}+\frac{3}{2}\right)\Rightarrow 3\mathrm{y}=3\left[\frac{2\mathrm{x}+3}{2}\right]\\ 6\mathrm{y}=6\mathrm{x}+9\Rightarrow 2\mathrm{x}-2\mathrm{y}+3=0\end{array}$
Slope of tangent = 1
Slope of normal = –1
Equation of normal at $(\frac{3}{2},3)$ is
$\begin{array}{l}\mathrm{y}-3=-1\left(\mathrm{x}-\frac{3}{2}\right)\Rightarrow \mathrm{y}-3=-\mathrm{x}+\frac{3}{2}\\ 2\mathrm{y}-6=-2\mathrm{x}+3\Rightarrow 2\mathrm{x}+2\mathrm{y}-9=0\end{array}$