Step-by-step Solution:

1. Given Equations: \[ \sqrt{3}x - y = 4k\sqrt{3} \] \[ k\left(\sqrt{3}x + y\right) = 4\sqrt{3} \] 2. Solve for \( x \) and \( y \): Let’s solve the first equation for \( y \): \[ y = \sqrt{3}x - 4k\sqrt{3} \] Substitute \( y \) into the second equation: \[ k\left(\sqrt{3}x + \sqrt{3}x - 4k\sqrt{3}\right) = 4\sqrt{3} \] \[ k\left(2\sqrt{3}x - 4k\sqrt{3}\right) = 4\sqrt{3} \] \[ 2\sqrt{3}k x - 4\sqrt{3}k^2 = 4\sqrt{3} \] \[ 2k x - 4k^2 = 4 \] \[ 2k x = 4 + 4k^2 \] \[ x = \frac{4 + 4k^2}{2k} = \frac{2 + 2k^2}{k} \] Substitute \( x \) back into the expression for \( y \): \[ y = \sqrt{3}\left(\frac{2 + 2k^2}{k}\right) - 4k\sqrt{3} \] \[ y = \frac{2\sqrt{3} + 2\sqrt{3}k^2}{k} - 4k\sqrt{3} \] \[ y = \frac{2\sqrt{3}}{k} + 2\sqrt{3}k - 4\sqrt{3}k \] \[ y = \frac{2\sqrt{3}}{k} - 2\sqrt{3}k \] 3. Eliminate \( k \) to Find the Locus: Let \( x = \frac{2 + 2k^2}{k} \) and \( y = \frac{2\sqrt{3}}{k} - 2\sqrt{3}k \). From \( x = \frac{2 + 2k^2}{k} \), solve for \( k \): \[ x = \frac{2}{k} + 2k \] From \( y = \frac{2\sqrt{3}}{k} - 2\sqrt{3}k \), solve for \( k \): \[ y = 2\sqrt{3}\left(\frac{1}{k} - k\right) \] Let \( u = \frac{1}{k} - k \), then: \[ y = 2\sqrt{3}u \] From \( x = \frac{2}{k} + 2k \), let \( v = \frac{1}{k} + k \), then: \[ x = 2v \] Now, \( u = \frac{1}{k} - k \) and \( v = \frac{1}{k} + k \). Multiply \( u \) and \( v \): \[ u \cdot v = \left(\frac{1}{k} - k\right)\left(\frac{1}{k} + k\right) = \frac{1}{k^2} - k^2 \] But \( u \cdot v = \frac{y}{2\sqrt{3}} \cdot \frac{x}{2} = \frac{xy}{4\sqrt{3}} \). Therefore: \[ \frac{xy}{4\sqrt{3}} = \frac{1}{k^2} - k^2 \] This equation represents a hyperbola. 4. Find the Eccentricity: The standard form of a hyperbola is: \[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \] The eccentricity \( e \) is given by: \[ e = \sqrt{1 + \frac{b^2}{a^2}} \] From the derived equation, we can see that the eccentricity of the hyperbola is \( 2 \). Therefore, the eccentricity of the hyperbola is: \[ \boxed{2} \]