If p and q are lengths of the perpendiculars | vedclass

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(B) 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

Vedclass · Accepted Answer

$4p^2 + q^2 = k^2$

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(B) 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 $x \sec 	heta + y \operatorname{cosec} 	heta - k = 0$,we have $p = \frac{|-k|}{\sqrt{\sec^2 	heta + \operatorname{cosec}^2 	heta}} = \frac{|k|}{\sqrt{\frac{1}{\cos^2 	heta} + \frac{1}{\sin^2 	heta}}} = \frac{|k| \sin 	heta \cos 	heta}{\sqrt{\sin^2 	heta + \cos^2 	heta}} = |k| \sin 	heta \cos 	heta = \frac{|k|}{2} \sin 2	heta$.Thus,$p^2 = \frac{k^2}{4} \sin^2 2	heta$.For the second line $x \cos 	heta - y \sin 	heta - k \cos 2	heta = 0$,we have $q = \frac{|-k \cos 2	heta|}{\sqrt{\cos^2 	heta + \sin^2 	heta}} = |k \cos 2	heta|$.Thus,$q^2 = k^2 \cos^2 2	heta$.Now,consider $4p^2 + q^2 = 4 \left( \frac{k^2}{4} \sin^2 2	heta ight) + k^2 \cos^2 2	heta = k^2 \sin^2 2	heta + k^2 \cos^2 2	heta = k^2 (\sin^2 2	heta + \cos^2 2	heta) = k^2$.Therefore,$4p^2 + q^2 = k^2$.

If $p$ and $q$ are lengths of the perpendiculars from the origin to the lines $x \sec \theta + y \operatorname{cosec} \theta = k$ and $x \cos \theta - y \sin \theta = k \cos 2\theta$ respectively,then

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