Question

$L_1$ and $L_2$ are $2$ tangent $y^2=16x$ which markes angle ${\tan }^{-1}\frac{1}{3}$ with the tangent ${}^{\prime }{T}^{\prime }$ to the same parabola at the end point of latus rectum which the ordinate then find the area of $△$ formed $L_1,L_2$ ?

Answer 1

$\begin{array}{c}y=16x\\ \therefore a=4\\ \left\{\left(2a{t}_{1}t{r}_{1}a\left(t_1+t_2\right)\right)\right\}\\ \left(-8,\frac{32}{3}\right)\\ \tan \theta =-t_1\\ -t_1=\tan {\tan }^{-1}\frac{1}{3}\\ -t_1=\frac{1}{3},t_1=\frac{-1}{3},\\ t_1{t}_2=-1,t_2=3\\ Then,areaoftriangle\left(T{T}_1{T}_2\right)is\\ \Rightarrow \frac{1}{2}\left[\frac{4}{9}\left(\frac{32}{3}-24\right)-8\left(24+\frac{8}{3}\right)+36\left(\frac{-8}{3}-\frac{32}{3}\right)\right]\\ \Rightarrow \frac{}{2}\times \left|\frac{4}{9}\times \frac{-40}{3}-8\times \frac{80}{3}-36\times \frac{40}{3}\right|\\ \Rightarrow \frac{1}{2}\times \left|\frac{160}{27}-\frac{640}{3}-\frac{1440}{3}\right|=\frac{1}{2}\left|\frac{180}{27}-693.33\right|\\ =\frac{1}{2}\left[5.92-693.33\right]=343.705squnit.\end{array}$