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

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$100$

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(D) The length of the perpendicular from the origin $(0, 0)$ to the line $Ax + By + C = 0$ is given by $d = \frac{|C|}{\sqrt{A^2 + B^2}}$.For the first line $x \sec \alpha + y \operatorname{cosec} \alpha - 10 = 0$,we have $p = \frac{|-10|}{\sqrt{\sec^2 \alpha + \operatorname{cosec}^2 \alpha}} = \frac{10}{\sqrt{\frac{1}{\cos^2 \alpha} + \frac{1}{\sin^2 \alpha}}} = \frac{10}{\sqrt{\frac{\sin^2 \alpha + \cos^2 \alpha}{\sin^2 \alpha \cos^2 \alpha}}} = 10 \sin \alpha \cos \alpha = 5 \sin 2 \alpha$.Thus,$p = 5 \sin 2 \alpha$,which implies $p^2 = 25 \sin^2 2 \alpha$,so $4p^2 = 100 \sin^2 2 \alpha$.For the second line $x \cos \alpha - y \sin \alpha - 10 \cos 2 \alpha = 0$,we have $q = \frac{|-10 \cos 2 \alpha|}{\sqrt{\cos^2 \alpha + \sin^2 \alpha}} = \frac{10 \cos 2 \alpha}{1} = 10 \cos 2 \alpha$.Thus,$q^2 = 100 \cos^2 2 \alpha$.Adding these,$4p^2 + q^2 = 100 \sin^2 2 \alpha + 100 \cos^2 2 \alpha = 100(\sin^2 2 \alpha + \cos^2 2 \alpha) = 100(1) = 100$.

If $p$ and $q$ are the lengths of the perpendiculars from the origin to the straight lines $x \sec \alpha + y \operatorname{cosec} \alpha = 10$ and $x \cos \alpha - y \sin \alpha = 10 \cos 2 \alpha$,then $4 p^2 + q^2 =$

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