The locus of the point of intersection of the lines

Question:

The locus of the point of intersection of the lines

$(\sqrt{3}) \mathrm{kx}+\mathrm{ky}-4 \sqrt{3}=0$ and

$\sqrt{3} x-y-4(\sqrt{3}) k=0$ is a conic, whose eccentricity is_________.

Solution:

$K=\frac{4 \sqrt{3}}{\sqrt{3} x+y}=\frac{\sqrt{3} x-y}{4 \sqrt{3}}$

$\Rightarrow 3 x^{2}-y^{2}=48$

$\Rightarrow \frac{x^{2}}{16}-\frac{y^{2}}{48}=1$

Now, $48=16\left(\mathrm{e}^{2}-1\right)$

$\Rightarrow \mathrm{e}=\sqrt{4}=2$

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