If p and q are the perpendicular distances from the origin to the straight lines $xsec\theta -y\cos ec\theta =a$ and $x\cos \theta +y\sin \theta =a\cos 2\theta$ then

Given lines $xsec\theta -y\cos ec\theta =a$

$$\begin{gathered} \Rightarrow \frac{x}{\cos \theta }-\frac{y}{\sin \theta }=a \\ \Rightarrow x\sin \theta -y\cos \theta -a\cos \theta \sin \theta =0---\left(1\right) \\ and x \cos \theta +y\sin \theta -a\cos 2\theta =0---\left(2\right) \end{gathered}$$

$P=\perp r$ distance from (0, 0) to (i)

$=\frac{\left|-a\cos \theta \sin \theta \right|}{\sqrt{{\cos }^2\theta +{\sin }^2\theta }}=a \cos \theta \sin \theta =\frac{a}{2}\sin 2\theta$

$q=\perp r$ distance from (0 0) to (ii)

$=\frac{\left|-a \cos 2\theta \right|}{\sqrt{{\cos }^2\theta +{\sin }^2\theta }}=a \cos 2\theta$

$4p^2+q^2=4\left(\frac{a^2}{4}{\sin }^{2}2\theta \right)+a^2{\cos }^{2}2\theta$

$$\begin{gathered} =a^2\left({\sin }^{2}2\theta +{\cos }^{2}2\theta \right) \\ =a^2\left(1\right) \\ =a^2 \end{gathered}$$