If p_{1} and p_{2} are the lengths of perpend | vedclass

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(C) The length of the perpendicular from the origin $(0,0)$ to the line $Ax + By + C = 0$ is given by $p = \frac{|C|}{\sqrt{A^2 + B^2}}$.For the first

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$25$

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(C) The length of the perpendicular from the origin $(0,0)$ to the line $Ax + By + C = 0$ is given by $p = \frac{|C|}{\sqrt{A^2 + B^2}}$.For the first line $x \sin 	heta + y \cos 	heta - 5 \cos 2 	heta = 0$,we have $p_{1} = \frac{|-5 \cos 2 	heta|}{\sqrt{\sin^2 	heta + \cos^2 	heta}} = |5 \cos 2 	heta|$.Thus,$p_{1}^2 = 25 \cos^2 2 	heta$.For the second line $x \operatorname{cosec} 	heta + y \sec 	heta - 5 = 0$,we have $p_{2} = \frac{|-5|}{\sqrt{\operatorname{cosec}^2 	heta + \sec^2 	heta}} = \frac{5}{\sqrt{\frac{1}{\sin^2 	heta} + \frac{1}{\cos^2 	heta}}} = \frac{5 \sin 	heta \cos 	heta}{\sqrt{\cos^2 	heta + \sin^2 	heta}} = 5 \sin 	heta \cos 	heta = \frac{5}{2} \sin 2 	heta$.Thus,$4 p_{2}^2 = 4 \left( \frac{25}{4} \sin^2 2 	heta ight) = 25 \sin^2 2 	heta$.Adding these,$p_{1}^2 + 4 p_{2}^2 = 25 \cos^2 2 	heta + 25 \sin^2 2 	heta = 25(\cos^2 2 	heta + \sin^2 2 	heta) = 25$.

If $p_{1}$ and $p_{2}$ are the lengths of perpendiculars from the origin to the lines $x \sin \theta + y \cos \theta = 5 \cos 2 \theta$ and $x \operatorname{cosec} \theta + y \sec \theta - 5 = 0$ respectively,then $p_{1}^{2} + 4 p_{2}^{2} = $

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