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(C) The distance $p$ of the origin $(0, 0)$ from the line $x \sec \alpha + y \csc \alpha - k = 0$ is given by $p = \frac{|-k|}{\sqrt{\sec^2 \alpha + \

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$k^2$

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(C) The distance $p$ of the origin $(0, 0)$ from the line $x \sec \alpha + y \csc \alpha - k = 0$ is given by $p = \frac{|-k|}{\sqrt{\sec^2 \alpha + \csc^2 \alpha}} = \frac{|k|}{\sqrt{\frac{1}{\cos^2 \alpha} + \frac{1}{\sin^2 \alpha}}} = |k \sin \alpha \cos \alpha|$.Thus,$p^2 = k^2 \sin^2 \alpha \cos^2 \alpha$.The distance $p'$ of the origin $(0, 0)$ from the line $x \cos \alpha - y \sin \alpha - k \cos 2\alpha = 0$ is given by $p' = \frac{|-k \cos 2\alpha|}{\sqrt{\cos^2 \alpha + \sin^2 \alpha}} = |k \cos 2\alpha|$.Thus,$p'^2 = k^2 \cos^2 2\alpha$.Now,$4p^2 + p'^2 = 4k^2 \sin^2 \alpha \cos^2 \alpha + k^2 \cos^2 2\alpha$.Using $2 \sin \alpha \cos \alpha = \sin 2\alpha$,we have $4 \sin^2 \alpha \cos^2 \alpha = \sin^2 2\alpha$.So,$4p^2 + p'^2 = k^2 \sin^2 2\alpha + k^2 \cos^2 2\alpha = k^2(\sin^2 2\alpha + \cos^2 2\alpha) = k^2$.

If $p$ and $p'$ are the distances of the origin from the lines $x \sec \alpha + y \csc \alpha = k$ and $x \cos \alpha - y \sin \alpha = k \cos 2\alpha$,then $4p^2 + p'^2$ equals:

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