The tangents to the parabola $y^2=4ax\text{ at }P\left(t_1\right)\text{ and }Q\left(t_2\right)$ intersect at R. The area of $∆PQR$ is 

$P\left(a{t}_1^2,2a{t}_1\right)Q\left(a{t}_2^2,2a{t}_2\right)R\left[a{t}_1{t}_2,a\left(t_1+t_2\right)\right]$

Area of the $\mathrm{\Delta }PQR=\frac{1}{2}\left|\begin{array}{cc}a\left(t_1^2-t_2^2\right)& 2a\left(t_1-t_2\right)\\ a\left(t_1^2-t_1{t}_2\right)& a\left(2t_1-t_1-t_2\right)\end{array}\right|=\frac{1}{2}\left|\begin{array}{l}a\left(t_1+t_2\right)\left(t_1-t_2\right)\hspace{0.25em} \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}2a\left(t_1-t_2\right)\\ a{t}_1\left(t_1-t_2\right)\hspace{0.25em} \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}a\left(t_1-t_2\right)\end{array}\right|$

$=\frac{1}{2}{a}^2{\left(t_1-t_2\right)}^2\left|\begin{array}{l}{t}_1+t_2 \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}2\\ t_1 \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\end{array}\right|=\frac{1}{2}{a}^2{\left(t_1-t_2\right)}^2\left|t_1+t_2-2t_1\right|=\frac{1}{2}{a}^2\left|{\left(t_1-t_2\right)}^3\right|$