If p_1 and p_2 are the lengths of the 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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$4 \csc^2 4\alpha$

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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 $x \sec \alpha + y \csc \alpha - 2a = 0$,we have $p_1 = \frac{|-2a|}{\sqrt{\sec^2 \alpha + \csc^2 \alpha}} = \frac{2|a|}{\sqrt{\frac{1}{\cos^2 \alpha} + \frac{1}{\sin^2 \alpha}}} = \frac{2|a| \sin \alpha \cos \alpha}{\sqrt{\sin^2 \alpha + \cos^2 \alpha}} = |a \sin 2\alpha|$.Thus,$p_1^2 = a^2 \sin^2 2\alpha$.For the second line $x \cos \alpha - y \sin \alpha - a \cos 2\alpha = 0$,we have $p_2 = \frac{|-a \cos 2\alpha|}{\sqrt{\cos^2 \alpha + \sin^2 \alpha}} = |a \cos 2\alpha|$.Thus,$p_2^2 = a^2 \cos^2 2\alpha$.Now,$\left( \frac{p_1}{p_2} + \frac{p_2}{p_1} \right)^2 = \left( \frac{p_1^2 + p_2^2}{p_1 p_2} \right)^2 = \frac{(a^2 \sin^2 2\alpha + a^2 \cos^2 2\alpha)^2}{a^2 \sin^2 2\alpha \cdot a^2 \cos^2 2\alpha} = \frac{(a^2)^2}{a^4 \sin^2 2\alpha \cos^2 2\alpha} = \frac{1}{\sin^2 2\alpha \cos^2 2\alpha}$.Multiplying numerator and denominator by $4$,we get $\frac{4}{4 \sin^2 2\alpha \cos^2 2\alpha} = \frac{4}{(2 \sin 2\alpha \cos 2\alpha)^2} = \frac{4}{\sin^2 4\alpha} = 4 \csc^2 4\alpha$.

If $p_1$ and $p_2$ are the lengths of the perpendiculars from the origin to the lines $x \sec \alpha + y \csc \alpha = 2a$ and $x \cos \alpha - y \sin \alpha = a \cos 2\alpha$ respectively,then what is the value of $\left( \frac{p_1}{p_2} + \frac{p_2}{p_1} \right)^2$?

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