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

$$\begin{gathered} \mathrm{Given} \mathrm{parabol} {\mathrm{y}}^2=4\mathrm{ax} \\ Let points on parabola be \\ {\mathrm{t}}_1=\left({{\mathrm{at}}_1}^2, 2{\mathrm{at}}_1\right)=\left({\mathrm{x}}_1, {\mathrm{y}}_1\right) \\ {\mathrm{t}}_2=\left({{\mathrm{at}}_2}^2, 2{\mathrm{at}}_2\right)=\left({\mathrm{x}}_2, {\mathrm{y}}_2\right) \\ {\mathrm{t}}_3=\left({{\mathrm{at}}_3}^2, 2{\mathrm{at}}_3\right)=\left({\mathrm{x}}_3, {\mathrm{y}}_3\right) \\ We know that area of triangle formed by \tan gents at the points t_1,t_2 and t_3 is \\ ∆=\frac{1}{16\mathrm{a}}\left|\left({\mathrm{y}}_1-{\mathrm{y}}_2\right) \left({\mathrm{y}}_2-{\mathrm{y}}_3\right) \left({\mathrm{y}}_3-{\mathrm{y}}_1\right)\right| \\ ∆=\frac{8{\mathrm{a}}^3}{16\mathrm{a}}\left|\left(t_1-t_2\right) \left(t_2-{\mathrm{t}}_3\right) \left({\mathrm{t}}_3-{\mathrm{t}}_1\right)\right| \\ ∆=\frac{{\mathrm{a}}^2}{2}\left|\left({\mathrm{y}}_1-{\mathrm{y}}_2\right) \left({\mathrm{y}}_2-{\mathrm{y}}_3\right) \left({\mathrm{t}}_3-{\mathrm{t}}_1\right)\right| \\ By comparision \\ \mathrm{k}=\frac{{\mathrm{a}}^2}{2} \end{gathered}$$