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(A) The first line is $x \operatorname{cosec} \alpha - y \sec \alpha = k \cot 2 \alpha$,which can be written as $\frac{x}{\sin \alpha} - \frac{y}{\cos

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$4 p^{2} + q^{2}$

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(A) The first line is $x \operatorname{cosec} \alpha - y \sec \alpha = k \cot 2 \alpha$,which can be written as $\frac{x}{\sin \alpha} - \frac{y}{\cos \alpha} = \frac{k \cos 2 \alpha}{\sin 2 \alpha}$.Multiplying by $\sin \alpha \cos \alpha$,we get $x \cos \alpha - y \sin \alpha = \frac{k \cos 2 \alpha}{\sin 2 \alpha} \cdot \sin \alpha \cos \alpha = \frac{k \cos 2 \alpha}{2 \sin \alpha \cos \alpha} \cdot \sin \alpha \cos \alpha = \frac{k}{2} \cos 2 \alpha$.The perpendicular distance $p$ from the origin $(0, 0)$ to the line $x \cos \alpha - y \sin \alpha - \frac{k}{2} \cos 2 \alpha = 0$ is $p = \left| \frac{-\frac{k}{2} \cos 2 \alpha}{\sqrt{\cos^{2} \alpha + \sin^{2} \alpha}} \right| = \left| \frac{k}{2} \cos 2 \alpha \right|$.Thus,$2p = |k \cos 2 \alpha|$,which implies $4p^{2} = k^{2} \cos^{2} 2 \alpha$ $(i)$.The second line is $x \sin \alpha + y \cos \alpha = k \sin 2 \alpha$. The perpendicular distance $q$ from the origin is $q = \left| \frac{-k \sin 2 \alpha}{\sqrt{\sin^{2} \alpha + \cos^{2} \alpha}} \right| = |k \sin 2 \alpha|$.Thus,$q^{2} = k^{2} \sin^{2} 2 \alpha$ $(ii)$.Adding $(i)$ and $(ii)$,we get $4p^{2} + q^{2} = k^{2} (\cos^{2} 2 \alpha + \sin^{2} 2 \alpha) = k^{2}$.

If $p$ and $q$ are the lengths of the perpendiculars from the origin on the lines,$x \operatorname{cosec} \alpha - y \sec \alpha = k \cot 2 \alpha$ and $x \sin \alpha + y \cos \alpha = k \sin 2 \alpha$ respectively,then $k^{2}$ is equal to :

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