The locus of intersection of the lines $\sqrt 3 x-y-4\sqrt3 t=0$ and $\sqrt 3 tx+ty-4\sqrt 3=0$(where t is a parameter) is a hyperbola . we have to find its eccentricity .
I got the intersection point as $x=(2+2t^2)/(t)$ and $y=(2\sqrt 3)/(2\sqrt 3 t^2 +t)$.
after that I got stuck .
The family of lines is
$$\begin{cases}&I\;\;\sqrt3\,x-y=4\sqrt3\,t\,/\,\cdot(-t)\implies-\sqrt3\,tx+ty=-4\sqrt3t^2\\{}\\ &II\;\;\sqrt3\,tx+ty=4\sqrt3\end{cases}\stackrel{\text{sum up}\,-tI+II}\implies$$
$$\;2ty=4\sqrt3(1-t^2)\implies \color{red}{y=\frac{2\sqrt3(1-t^2)}t}\stackrel{\text{substitute in I}}\implies$$
$$\sqrt3\,x=4\sqrt3\,t+\frac{2\sqrt3(1-t^2)}t\implies$$
$$ \color{red}{x=4t+\frac{2(1-t^2)}t=\frac{2(t^2+1)}t}$$
so
$$\begin{cases}&x^2=\cfrac{4(t^2+1)^2}{t^2}\\{}\\ &y^2=\cfrac{12(t^2-1)^2}{t^2}\end{cases}\implies 3x^2-y^2=48\implies\cfrac{x^2}{4^2}-\cfrac{y^2}{(4\sqrt3)^2}=1$$
and thus the eccentricity is
$$e=\frac{\sqrt{a^2+b^2}}a=\frac{\sqrt{16+48}}{4}=\frac{\sqrt{64}}{4}=2$$