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 $\mathrm{k}^{2}$ is equal to :
Correct Option: 1
First line is $\frac{\mathrm{x}}{\sin \alpha}-\frac{\mathrm{y}}{\cos \alpha}=\frac{\mathrm{k} \cos 2 \alpha}{\sin 2 \alpha}$
$\Rightarrow x \cos \alpha-y \sin \alpha=\frac{k}{2} \cos 2 \alpha$
$\Rightarrow \mathrm{p}=\left|\frac{\mathrm{k}}{2} \cos \alpha\right| \Rightarrow 2 \mathrm{p}=|\mathrm{k} \cos 2 \alpha|$......(i)
second line is $\mathrm{x} \sin \alpha+\mathrm{y} \cos \alpha=\mathrm{k} \sin 2 \alpha$
$\Rightarrow \mathrm{q}=|\mathrm{ksin} 2 \alpha|$.........(ii)
Hence $4 p^{2}+q^{2}=k^{2} \quad($ From (i) & (ii))
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