At a point P on the parabola ${\mathrm{y}}^2=4\mathrm{ax}(\mathrm{a}>0)$ tangent and normal are drawn. Tangent intersect x-axis at Q and normal intersect the parabola at R such that chord PR – subtends 90° at vertex then 

Let $\mathrm{P}({\mathrm{at}}_1^2,2{\mathrm{at}}_1)$
Equation of tangent at P is ${\mathrm{t}}_1\mathrm{y}=\mathrm{x}+{\mathrm{at}}_1^2$
On X-axis y = 0 so $\mathrm{Q}=(-{\mathrm{at}}_1^2,0)$
Normal at P intersect the parabola at $\mathrm{R}({\mathrm{at}}_2^2,2{\mathrm{at}}_2)\Rightarrow {\mathrm{t}}_2=-{\mathrm{t}}_1-\frac{2}{{\mathrm{t}}_1}$….(1)
Slope of OP is ${\mathrm{m}}_1=\frac{2}{{\mathrm{t}}_1}$
Slope of OR is ${\mathrm{m}}_2=\frac{2}{{\mathrm{t}}_2}$
$$\begin{gathered} \mathrm{OP}\perp \mathrm{OR}\Rightarrow {\mathrm{t}}_1{\mathrm{t}}_2=-4\Rightarrow {\mathrm{t}}_2=\frac{-4}{{\mathrm{t}}_1} \\ \left(1\right)\Rightarrow \frac{-4}{{\mathrm{t}}_1}=-{\mathrm{t}}_1-\frac{2}{{\mathrm{t}}_1} \\ \frac{2}{{\mathrm{t}}_1}={\mathrm{t}}_1\Rightarrow {\mathrm{t}}_1^2=2\Rightarrow {\mathrm{t}}_1=\pm \sqrt{2} \\ {\mathrm{t}}_2=\mp \sqrt{2}\Rightarrow \mathrm{PQ}=2{\mathrm{at}}_1\sqrt{{\mathrm{t}}_1^2+1}=2\mathrm{a}\sqrt{6} \\ \mathrm{P}=(2\mathrm{a},2\sqrt{2}\mathrm{a}) \mathrm{Q}=(-2\mathrm{a},0) \end{gathered}$$  
Area of $\mathrm{\Delta PQR}=\frac{1}{2}\left|\begin{array}{ccc}2\mathrm{a}& 2\sqrt{2}\mathrm{a}& 1\\ -2\mathrm{a}& 0& 1\\ 8\mathrm{a}& -4\sqrt{2}\mathrm{a}& 1\end{array}\right|=18\sqrt{2}{\mathrm{a}}^2$