The area of triangle formed by the points $t_1,t_2$ and $t_3$ on $y^2=4ax$ is $k\left|(t_1-t_2)(t_2-t_3)(t_3-t_1)\right|$ then $K=$

$\text{ Area }=\frac{1}{2}\left|\begin{array}{l}a{t}_1^2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}2a{t}_1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\\ a{t}_2^2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}2a{t}_2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\\ a{t}_3^2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}2a{t}_3\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\end{array}\right|$

$=\frac{2a^2}{2}\left|\begin{array}{ccc}1& t_1& {t_1}^2\\ 1& t_2& {t_2}^2\\ 1& t_3& {t_3}^2\end{array}\right|$

$=a^2\left|\left(t_1-t_2\right)\left(t_2-t_3\right)\left(t_3-t_1\right)\right|$