Question:
The locus of the point of intersection 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
Solution:
$\sqrt{3} \mathrm{kx}+\mathrm{ky}=4 \sqrt{3}$
$\sqrt{3} \mathrm{kx}-\mathrm{ky}=4 \sqrt{3} \mathrm{k}^{2}$
Adding equation (1) $\backslash \&(2) 2 \sqrt{3} \mathrm{kx}=4 \sqrt{3}\left(\mathrm{k}^{2}+1\right)$
$x=2\left(k+\frac{1}{k}\right)$
Substracting equation (1) $\backslash \&(2) y=2 \sqrt{3}\left(\frac{1}{k}-k\right)$
$\therefore \frac{\mathrm{x}^{2}}{4}-\frac{\mathrm{y}^{2}}{12}=4$
$\frac{x^{2}}{16}-\frac{y^{2}}{48}=1 \quad$ Hyperbola
$\therefore \mathrm{e}^{2}=1+\frac{48}{16}$
$e=2$