If $p$ and $q$ are the lengths of perpendiculars from the origin to the lines $x \cos \theta-y \sin \theta=k \cos 2 \theta$ and $x \sec \theta+y \operatorname{cosec} \theta=k$, respectively, prove that $p^{2}+4 q^{2}=k^{2}$
The equations of given lines are
$x \cos \theta-y \sin \theta=k \cos 2 \theta \ldots$ (1)
$x \sec \theta+y \operatorname{cosec} \theta=k$ (2)
The perpendicular distance $(d)$ of a line $A x+B y+C=0$ from a point $\left(x_{1}, y_{1}\right)$ is given by $d=\frac{\left|A x_{1}+B y_{1}+C\right|}{\sqrt{A^{2}+B^{2}}}$.
On comparing equation (1) to the general equation of line i.e., $A x+B y+C=0$, we obtain $A=\cos \theta, B=-\sin \theta$, and $C=-k \cos 2 \theta$.
It is given that p is the length of the perpendicular from (0, 0) to line (1).
$\therefore p=\frac{|A(0)+B(0)+C|}{\sqrt{A^{2}+B^{2}}}=\frac{|C|}{\sqrt{A^{2}+B^{2}}}=\frac{|-k \cos 2 \theta|}{\sqrt{\cos ^{2} \theta+\sin ^{2} \theta}}=|-k \cos 2 \theta|$ (3)
On comparing equation (2) to the general equation of line i.e., $A x+B y+C=0$, we obtain $A=\sec \theta, B=\operatorname{cosec} \theta$, and $C=-k$.
It is given that q is the length of the perpendicular from (0, 0) to line (2).
$\therefore q=\frac{|A(0)+B(0)+C|}{\sqrt{A^{2}+B^{2}}}=\frac{|C|}{\sqrt{A^{2}+B^{2}}}=\frac{|-k|}{\sqrt{\sec ^{2} \theta+\operatorname{cosec}^{2} \theta}}$ (4)
From $(3)$ and $(4)$, we have
$p^{2}+4 q^{2}=(|-k \cos 2 \theta|)^{2}+4\left(\frac{|-k|}{\sqrt{\sec ^{2} \theta+\operatorname{cosec}^{2} \theta}}\right)^{2}$
$=k^{2} \cos ^{2} 2 \theta+\frac{4 k^{2}}{\left(\sec ^{2} \theta+\operatorname{cosec}^{2} \theta\right)}$
$=k^{2} \cos ^{2} 2 \theta+\frac{4 k^{2}}{\left(\frac{1}{\cos ^{2} \theta}+\frac{1}{\sin ^{2} \theta}\right)}$
$=k^{2} \cos ^{2} 2 \theta+\frac{4 k^{2}}{\left(\frac{\sin ^{2} \theta+\cos ^{2} \theta}{\sin ^{2} \theta \cos ^{2} \theta}\right)}$
$=k^{2} \cos ^{2} 2 \theta+\frac{4 k^{2}}{\left(\frac{1}{\sin ^{2} \theta \cos ^{2} \theta}\right)}$
$=k^{2} \cos ^{2} 2 \theta+4 k^{2} \sin ^{2} \theta \cos ^{2} \theta$
$=k^{2} \cos ^{2} 2 \theta+k^{2}(2 \sin \theta \cos \theta)^{2}$
$=k^{2} \cos ^{2} 2 \theta+k^{2} \sin ^{2} 2 \theta$
$=k^{2}\left(\cos ^{2} 2 \theta+\sin ^{2} 2 \theta\right)$
$=k^{2}$
Hence, we proved that $p^{2}+4 q^{2}=k^{2}$.