The area of triangle formed by the points ${\mathrm{t}}_1,{\mathrm{t}}_2$and ${\mathrm{t}}_3$ on ${\mathrm{y}}^2=4\mathrm{ax}$ is $\mathrm{kl}({\mathrm{t}}_1-{\mathrm{t}}_2) ({\mathrm{t}}_2-{\mathrm{t}}_3) \overline{) ({\mathrm{t}}_3-{\mathrm{t}}_1)}$  then K =

$$\begin{gathered} {\mathrm{y}}^2=4\mathrm{ax} \\ {\mathrm{t}}_1=({{\mathrm{at}}_1}^2, 2{\mathrm{at}}_1)=({\mathrm{x}}_1, {\mathrm{y}}_1) \\ {\mathrm{t}}_2=({{\mathrm{at}}_2}^2, 2{\mathrm{at}}_2)=({\mathrm{x}}_2, {\mathrm{y}}_2) \\ {\mathrm{t}}_3=({{\mathrm{at}}_3}^2, 2{\mathrm{at}}_3)=({\mathrm{x}}_3, {\mathrm{y}}_3) \\ We know that area of triangle formed by the vertices t_1,t_2,t_3 is \\ ∆=\frac{1}{8\mathrm{a}}\left|\left({\mathrm{y}}_1-{\mathrm{y}}_2\right) \left({\mathrm{y}}_2-{\mathrm{y}}_3\right) \left({\mathrm{y}}_3-{\mathrm{y}}_1\right)\right| \\ =\frac{1}{8\mathrm{a}}\left|\left(2{\mathrm{at}}_1-2{\mathrm{at}}_2\right) \left(2{\mathrm{at}}_2-2{\mathrm{at}}_3\right) \left(2{\mathrm{at}}_3-2{\mathrm{at}}_1\right)\right| \\ =\frac{8{\mathrm{a}}^3}{8\mathrm{a}}\left|\left({\mathrm{t}}_1-{\mathrm{t}}_2\right) \left({\mathrm{t}}_2-{\mathrm{t}}_3\right) \left({\mathrm{t}}_3-{\mathrm{t}}_1\right)\right| \\ By comparision \\ \Rightarrow \mathrm{k}={\mathrm{a}}^2 \end{gathered}$$