Tangent and normal are drawn at the upper and $\left({\mathrm{x}}_1{\mathrm{y}}_1\right)$ of the Latus rectum P with ${\mathrm{x}}_1>0, {\mathrm{y}}_1>0$ of the hyperbola $\frac{{\mathrm{x}}^2}{4}-\frac{{\mathrm{y}}^2}{12}=1$ intersecting the transverse axis at T and G respectively. Then $\frac{1}{9}$ (area of $\mathrm{△}\mathrm{PTG}$ ) 

$$\begin{gathered} Given hyperbola \frac{x^2}{4}-\frac{y^2}{12}=1 \\ where a^2=4,b^2=12 and e=2 \end{gathered}$$

$\mathrm{P}\left(\mathrm{ae}, {\mathrm{b}}^2/\mathrm{a}\right)=(4,6)$be positive end point of latusrectum

 Equation of tangent  at P is $\frac{\mathrm{ex}}{\mathrm{a}}-\frac{\mathrm{y}}{\mathrm{a}}=1$

Put $\mathrm{y}=0\Rightarrow \mathrm{T}=\left(\frac{\mathrm{a}}{\mathrm{e}}, 0\right)=(1,0)$

Equation of normal at P is

$$\begin{gathered} \frac{a^{2}x}{ae}+\frac{b^{2}y}{{\displaystyle \frac{b^2}{a}}}=a^2+b^2 \\ \Rightarrow put y=0 then \\ x=a{e}^3 \end{gathered}$$

$$\begin{gathered} Normal meets x-axis at G\Rightarrow \mathrm{G}=\left({\mathrm{ae}}^3,0\right)=(16,0) \\ \therefore \text{ area of }\mathrm{\Delta PTG}=\frac{1}{2}\left|4\left(0\right)+1(-6)+16\left(6\right)\right|=45 \\ Now \frac{1}{9}(\mathrm{area} \mathrm{of} \mathrm{\Delta PTG})=5 \end{gathered}$$