If $p_{1}$ and $p_{2}$ are the lengths of perendiculars from the origin to the lines $x \sin heta+y \cos heta=5 \cos 2 heta$ and $x \operatorname{cosec} heta+y \sec heta-5=0$ respectively, then $p_{1}^{2}+4 p_{2}^{2}=$

Question

If $p_{1}$ and $p_{2}$ are the lengths of perendiculars from the origin to the lines $x \sin 	heta+y \cos 	heta=5 \cos 2 	heta$ and $x \operatorname{cosec} 	heta+y \sec 	heta-5=0$ respectively, then $p_{1}^{2}+4 p_{2}^{2}=$

Aicharya · Accepted Answer

As per condition given, we write $\begin{aligned} \mathrm{p}_{1} &=\frac{|-5 \cos 2 	heta|}{\sqrt{\sin ^{2} 	heta+\cos ^{2} 	heta}} 	ext { and } p_{2}=\frac{|-5|}{\sqrt{\operatorname{cosec}^{2} 	heta+\sec ^{2} 	heta}} \ 	herefore \mathrm{p}_{1}^{2}=\frac{25 \cos ^{2} 2 	heta}{1} 	ext { and } 4 \mathrm{p}_{2}^{2} &=\frac{4(25)}{\operatorname{cosec}^{2} 	heta+\sec ^{2} 	heta} \ 	herefore \mathrm{p}_{1}^{2}+4 \mathrm{p}_{2}^{2} &=25\left(\cos ^{2} 	heta-\sin ^{2} 	hetaight)^{2}+\frac{100}{\left(\frac{1}{\sin ^{2} 	heta}ight)+\frac{1}{\cos ^{2} 	heta}} \ &=25(\cos 	heta-\sin 	heta)^{2}(\cos 	heta+\sin 	heta)^{2}+100 \sin ^{2} 	heta \cos ^{2} 	heta \ &=25(1-\sin 2 	heta)(1+\sin 2 	heta)+25\left(4 \sin ^{2} 	heta \cos ^{2} 	hetaight) \ &=25\left(1-\sin 2 	heta+\sin 2 	heta-\sin ^{2} 2 	hetaight)+25(2 \sin 	heta \cos 	heta)^{2} \ &=25\left(1-\sin ^{2} 2 	hetaight)+25(\sin 2 	heta)^{2} \ &=25\left(\cos ^{2} 2 	hetaight)+25\left(\sin ^{2} 2 	hetaight)=25 \end{aligned}$

If $p_{1}$ and $p_{2}$ are the lengths of perendiculars from the origin to the lines $x \sin \theta+y \cos…

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