If p and q are the length of perpendicular from the origin to the lines xcos θ−ysin θ=kcos2θ and xsec θ+ycosec θ=k respectively prove that p^2 +4q^2 =k^2


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Question Number 51822 by peter frank last updated on 30/Dec/18

$I f p a n d q a r e t h e l e n g t h$ $o f p e r p e n d i c u l a r f r o m$ $t h e o r i g i n t o t h e l i n e s$ $x \cos θ - y \sin θ = k c o s 2 θ$ $a n d x \sec θ + y cosec θ = k$ $r e s p e c t i v e l y$ $p r o v e t h a t$ $p^{2} + 4 q^{2} = k^{2}$
	Answered by tanmay.chaudhury50@gmail.com last updated on 30/Dec/18

	$p =∣ \frac{0 + 0 - k}{\sqrt{{c o s}^{2} θ + {s i n}^{2} θ}} ∣= k$ $q =∣ \frac{- k}{\sqrt{{s e c}^{2} θ + {c o s e c}^{2} θ}} ∣= \frac{k}{\sqrt{\frac{1}{{c o s}^{2} θ} + \frac{1}{{s i n}^{2} θ}}} = k s i n θ c o s θ$ $p l s c h e c k t h e q u e s t i o n . . .$
	Commented by peter frank last updated on 30/Dec/18

	$n o w f i x e d$
	Commented by peter frank last updated on 30/Dec/18

	$p^{2} = k^{2} \cos^{2} 2 θ . . . . (i)$ $q^{2} = \frac{1}{4} k^{2} \sin^{2} 2 θ . . . . (i i)$ $(i) + (i i)$ $p^{2} + 4 q^{2} = k^{2}$ $t h a n k y o u s i r f o r y o u r m e t h o d . . . . . f i n a l y i r e a c h$ $d e s t i n a t i o n$
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