The perpendicular distance from the origin to the line $\mathrm{x}$ $\cos \theta-\mathrm{y} \sin \theta=\mathrm{k} \cos 2 \theta$...(i)
$\mathrm{p}=\frac{\mathrm{k} \cos 2 \theta}{\sqrt{\cos ^2 \theta+\sin ^2 \theta}}=\mathrm{k} \cos 2 \theta,$
the other line is $x \sec \theta+y \operatorname{cosec} \theta=k$
or $\frac{x}{\cos \theta}+\frac{y}{\sin \theta}=k$
or $x \sin \theta+y \cos \theta=k \sin \theta \cos \theta$
or $x \sin \theta+y \cos \theta=\frac{k}{2} \sin 2 \theta \quad$... (ii)
$\therefore$ the perpendicular distance $q$ from the origin to the line(ii)
$\begin{gathered}
q=\frac{\frac{k}{2} \sin 2 \theta}{\sqrt{\sin ^2 \theta+\cos ^2 \theta}}=\frac{k}{2} \sin 2 \theta \\
\text { Now } p^2+4 q^2=k^2 \cos ^2 2 \theta+4\left(\frac{k}{2} \sin 2 \theta\right)^2 \\
\Rightarrow \quad p^2+4 q^2=k^2 \cos ^2 2 \theta+4 \cdot \frac{k^2}{4} \sin ^2 2 \theta \\
p^2+4 q^2=k^2\left(\cos ^2 2 \theta+\sin ^2 2 \theta\right)
\end{gathered}$
Hence, $p^2+4 q^2=k^2$