If p and q are the lengths of the perpendiculars from the origin on the lines, x cosec α - y sec α…

$L_1:xcosec\alpha -y\sec \alpha =k\cot 2\alpha$
$\frac{x}{\sin \alpha }-\frac{y}{\cos \alpha }=k\frac{\cos 2\alpha }{\sin 2\alpha }$

$\frac{x\cos \alpha -y\sin \alpha }{\sin \alpha \cos \alpha }=k\frac{\cos 2\alpha }{2 \sin \alpha \cos \alpha }$
$\Rightarrow x\cos \alpha -y\sin \alpha =\frac{k}{2}\cos 2\alpha$

We know that Perpendicular distance from a point $\left(x_1, y_1\right)$to $ax +by + c = 0$ is $\left|\frac{a{x}_1 + b{y}_1 + c}{\sqrt{a^2+b^2}}\right|$

Perpendicular distance from $(0,0)$ is $p=\left|\frac{0-0-\frac{k}{2}\cos 2\alpha }{\sqrt{{\cos }^2\alpha +{\sin }^2\alpha }}\right|$

$\Rightarrow p^2=\frac{k^2}{4}{\cos }^{2}2\alpha$

$L_2:x\sin \alpha +y\cos \alpha -k\sin 2\alpha =0$

Perpendicular distance from $(0,0)$ is $q=\left|\frac{0+0-k\sin 2\alpha }{\sqrt{{\sin }^2\alpha +{\cos }^2\alpha }}\right|$

$\Rightarrow q^2=k^2{\sin }^{2}2\alpha$

$4p^2+q^2=k^2\left({\sin }^{2}2\alpha +{\cos }^{2}2\alpha \right)$

We know that ${\sin }^{2}t + {\cos }^{2}t = 1.$
$4p^2+q^2=k^2$