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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