Q.The locus of the point of intersection of lines $\sqrt{3}x - y - 4\sqrt{3}k = 0$ and $\sqrt{3}kx + ky - 4\sqrt{3} = 0$ for different value of $k$ is a hyperbola whose eccentricity is 2.

Concept: Locus obtained by eliminating a parameter; identifying the conic.

From the first line, $\sqrt{3}x - y = 4\sqrt{3}\,k$. From the second, $k(\sqrt{3}x + y) = 4\sqrt{3}$, so $\sqrt{3}x + y = \dfrac{4\sqrt{3}}{k}$.

Multiply the two to cancel $k$:

This is a hyperbola with $a^2 = 16$, $b^2 = 48$, so …

The two given lines each depend on a parameter $k$; for each value of $k$ they meet at a point $(x, y)$. As $k$ varies, that intersection point moves and traces a curve — the locus. To find it we eliminate $k$ between the two equations, leaving a relation in $x$ and $y$ only.

The lines are

Step 1 — Isolate the $k$-dependence in each line

From $L_1$:

From $L_2$, factor $k$ out of the first two terms:

Step 2 — Multiply the two relations to cancel $k$

Multiplying (1) by (2), the $k$ on the right cancels:

The left side is a difference of squares:

Step 3 — Put the locus in standard hyperbola form

Divide through by $48$:

This is a hyperbola $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$ with

Step 4 — Compute the eccentricity …