Question

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

Answer 1

Answer: 2

$K=\frac{4\sqrt{3}}{\sqrt{3}x+y}=\frac{\sqrt{3}x-y}{4\sqrt{3}}$
$\Rightarrow 3x^2-y^2=48$
$\Rightarrow \frac{x^2}{16}-\frac{y^2}{48}=1$
Now, $48=16\left(e^2-1\right)$
$\Rightarrow e=\sqrt{4}=2$