The locus of the point of intersetion of the lines $\sqrt{3}\mathrm{kx}+\mathrm{ky}-4\sqrt{3}=0$ and $\sqrt{3}\mathrm{x}-\mathrm{y}-4\sqrt{3}\mathrm{k}=0$ is a conic whose eccentricity is $\mathrm{\_\_\_\_\_}$

$$\begin{gathered} \mathrm{Given} \mathrm{lines} \mathrm{are} \mathrm{K}(\sqrt{3}\mathrm{x}+\mathrm{y})=4\sqrt{3} \\ \Rightarrow \mathrm{K}=\frac{4\sqrt{3}}{\sqrt{3}\mathrm{x}+\mathrm{y}} \\ and \sqrt{3}\mathrm{x}-\mathrm{y}=4\sqrt{3}\mathrm{k} \\ \Rightarrow \sqrt{3}\mathrm{x}-\mathrm{y}=4\sqrt{3}\left(\frac{4\sqrt{3}}{\sqrt{3}\mathrm{x}+\mathrm{y}}\right) \\ \Rightarrow 3{\mathrm{x}}^2-{\mathrm{y}}^2=48 \\ \Rightarrow \frac{{\mathrm{x}}^2}{16}-\frac{{\mathrm{y}}^2}{48}=1 \\ Here conic is hyperbola \\ \therefore \mathrm{e}=\sqrt{\frac{16+48}{16}}=\sqrt{\frac{64}{16}}=\sqrt{4}=2 \end{gathered}$$