If a normal chord drawn at $\text{'}\mathrm{t}\text{'}$ on ${\mathrm{y}}^2=4\mathrm{ax}$ subtends a right angle at the focus then ${\mathrm{t}}^2=$

$$\begin{gathered} A normal to the parabola at t=t_1 intersects the parabola again at t_2 then \\ t_2=-t_1-\frac{2}{t_1} \\ Thus, the two points are A=(a{t_1}^2,2a{t}_1) and B=(a{t_2}^2,2a{t}_2) \\ slopes of lines AS and BS are \frac{2t_1}{{t^2}_1-1} and \frac{2t_2}{{t^2}_2-1} \\ Since they make a right angle at the focus , i.e. (a,0), \\ we get \left(\frac{2t_1}{{t^2}_1-1}\right(\frac{2t_2}{{t^2}_2-1})=-1 \end{gathered}$$

$$\begin{gathered} {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_{\mathrm{at} t} =\frac{2\mathrm{a}}{2\mathrm{at}}=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\mathrm{t}$}\right. \\ \mathrm{Slope} \mathrm{of} \mathrm{Normal} =-\mathrm{t} \\ \mathrm{Focus} =(a,0) \\ \mathrm{Slope} =\frac{0-2\mathrm{at}}{\mathrm{a}-{\mathrm{at}}^2}=\frac{-2\mathrm{at}}{\mathrm{a}(1-{\mathrm{t}}^2)} \\ \mathrm{Slope} \mathrm{of} \mathrm{Normal} =\frac{1-{\mathrm{t}}^2}{2\mathrm{t}} \end{gathered}$$

$$\begin{gathered} \mathrm{By} \mathrm{conceptually} \\ \mathrm{subtent} \mathrm{Right} \mathrm{angle} \\ \mathrm{at} \mathrm{focus} \mathrm{Then} {\mathrm{t}}^2=4 \end{gathered}$$