Parabola

Given, parabola is $y^2=x$ and $t_1=2 t_2=-4$ and $t_3=6$.So, intersection of tangents at $t_1$ and $t_2$, then tangents are $2y=x+1$ and $-4y=x+4$Solving these two equations, we get $y = \left(\frac{-1}{2}\right)$ and $x=-2$So, point $L = \left(-2, -\frac{1}{2}\right)$Now, intersection of tangents at $t_2=-4$ and $t_3=6$, then tangents are$-4y=x+4, 6y=x+9$Solving these two,we get $x=-6, y=\frac{1}{2}$So, print $M = \left(-6, \frac{1}{2}\right)$Now, intersection of tangents at $t_1=2$ and $t_3=6$, so tangents are $2y=x+1$ and $6y=x+9$Solving these two equation, we get $x=3, y=2$So, point $N=(3,2)$Area of triangle with vertices $L \left(-2, -\frac{1}{2}\right), M \left(-6, \frac{1}{2}\right)$ and $N(3,2)$ is$\; \text{Area} \; \; = \left| \frac{1}{2} \right| \begin{array}{ccc} -2 & -\frac{1}{2} & 1 \end{array}$$-6 \; \; \frac{1}{2} \; 1 | |$$3 \; 2 \; 1$$\; = \left| \frac{-15}{2} \right| = 7.5$