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(A) The given equations of the straight lines are:$(i) \ x \sec 	heta - y \operatorname{cosec} 	heta = a$$(ii) \ x \cos 	heta + y \sin 	heta = a \

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$4 p^2 + q^2 = a^2$

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(A) The given equations of the straight lines are:$(i) \ x \sec 	heta - y \operatorname{cosec} 	heta = a$$(ii) \ x \cos 	heta + y \sin 	heta = a \cos 2 	heta$The perpendicular distance $d$ from the origin $(0, 0)$ to a line $Ax + By + C = 0$ is given by $d = \frac{|C|}{\sqrt{A^2 + B^2}}$.For line $(i)$:$p = \frac{|-a|}{\sqrt{\sec^2 	heta + \operatorname{cosec}^2 	heta}} = \frac{a}{\sqrt{\frac{1}{\cos^2 	heta} + \frac{1}{\sin^2 	heta}}} = \frac{a}{\sqrt{\frac{\sin^2 	heta + \cos^2 	heta}{\sin^2 	heta \cos^2 	heta}}} = a \sin 	heta \cos 	heta = \frac{a}{2} \sin 2 	heta$.Thus,$2p = a \sin 2 	heta$.For line $(ii)$:$q = \frac{|-a \cos 2 	heta|}{\sqrt{\cos^2 	heta + \sin^2 	heta}} = \frac{a \cos 2 	heta}{1} = a \cos 2 	heta$.Now,calculating $4p^2 + q^2$:$4p^2 + q^2 = (2p)^2 + q^2 = (a \sin 2 	heta)^2 + (a \cos 2 	heta)^2$$= a^2 \sin^2 2 	heta + a^2 \cos^2 2 	heta = a^2 (\sin^2 2 	heta + \cos^2 2 	heta) = a^2(1) = a^2$.Therefore,$4p^2 + q^2 = a^2$.

If $p$ and $q$ are the perpendicular distances from the origin to the straight lines $x \sec \theta - y \operatorname{cosec} \theta = a$ and $x \cos \theta + y \sin \theta = a \cos 2 \theta$,then

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