Answer : True
The given equations are $\sqrt 3x-y-4\sqrt 3k=0$-----(1) and $\sqrt 3kx+ky-4\sqrt 3=0$----(2)
From equ(1) we get
$k=\large\frac{\sqrt 3x-y}{4\sqrt 3}$
Substituting the value of $'k'$ in equ(2) we get,
$\sqrt 3x\big(\large\frac{\sqrt 3x-y}{4\sqrt 3}\big)+\big(\large\frac{\sqrt 3x-y}{4\sqrt 3}\big)$$y=4\sqrt 3$
On simplifying we get,
$3x^2-y^2=48$
$\Rightarrow \large\frac{x^2}{16}-\frac{y^2}{48}$$=1$
This is clearly an equation of a hyperbola.
Hence the statement is True.