We have rectangular hyperbola

$x^2-y^2=a^2$

Given $ABC$ is an equilateral triangle.

$\therefore$$AB=BC=AC$

$A{B}^2=B{C}^2$

$a^2\left(\sec \theta +1\right){}^2+a^2{\tan }^2\theta =4a^2{\tan }^2\theta$

$\Rightarrow \left(\sec \theta +1\right){}^2=3{\tan }^2\theta$

$\Rightarrow \left(\sec \theta +1\right){}^2=3\left({\sec }^2\theta -1\right)$

$\Rightarrow \left(\sec \theta +1\right){}^2=3\left(\sec \theta +1\right)\left(\sec \theta -1\right)$

$\Rightarrow \sec \theta +1=3\sec \theta -3$

$\Rightarrow \sec \theta =2$

$\Rightarrow \theta =60°$

$\because$ Side $BC=2a\tan \theta$

$=2a\tan 60°=2a\sqrt{3}$

But side of triangle is $ka$.

$\therefore$$ka=2a\sqrt{3}$

$k=2\sqrt{3}$

Hence, $k\in (2,4]$