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If $p_{1}$ and $p_{2}$ are the lengths of perendiculars from the origin to the lines
$x \sin \theta+y \cos \theta=5 \cos 2 \theta$ and $x \operatorname{cosec} \theta+y \sec \theta-5=0$ respectively, then
$p_{1}^{2}+4 p_{2}^{2}=$
Options:
$x \sin \theta+y \cos \theta=5 \cos 2 \theta$ and $x \operatorname{cosec} \theta+y \sec \theta-5=0$ respectively, then
$p_{1}^{2}+4 p_{2}^{2}=$
Solution:
2076 Upvotes
Verified Answer
The correct answer is:
25
As per condition given, we write
$$
\begin{aligned}
\mathrm{p}_{1} &=\frac{|-5 \cos 2 \theta|}{\sqrt{\sin ^{2} \theta+\cos ^{2} \theta}} \text { and } p_{2}=\frac{|-5|}{\sqrt{\operatorname{cosec}^{2} \theta+\sec ^{2} \theta}} \\
\therefore \mathrm{p}_{1}^{2}=\frac{25 \cos ^{2} 2 \theta}{1} \text { and } 4 \mathrm{p}_{2}^{2} &=\frac{4(25)}{\operatorname{cosec}^{2} \theta+\sec ^{2} \theta} \\
\therefore \mathrm{p}_{1}^{2}+4 \mathrm{p}_{2}^{2} &=25\left(\cos ^{2} \theta-\sin ^{2} \theta\right)^{2}+\frac{100}{\left(\frac{1}{\sin ^{2} \theta}\right)+\frac{1}{\cos ^{2} \theta}} \\
&=25(\cos \theta-\sin \theta)^{2}(\cos \theta+\sin \theta)^{2}+100 \sin ^{2} \theta \cos ^{2} \theta \\
&=25(1-\sin 2 \theta)(1+\sin 2 \theta)+25\left(4 \sin ^{2} \theta \cos ^{2} \theta\right) \\
&=25\left(1-\sin 2 \theta+\sin 2 \theta-\sin ^{2} 2 \theta\right)+25(2 \sin \theta \cos \theta)^{2} \\
&=25\left(1-\sin ^{2} 2 \theta\right)+25(\sin 2 \theta)^{2} \\
&=25\left(\cos ^{2} 2 \theta\right)+25\left(\sin ^{2} 2 \theta\right)=25
\end{aligned}
$$
$$
\begin{aligned}
\mathrm{p}_{1} &=\frac{|-5 \cos 2 \theta|}{\sqrt{\sin ^{2} \theta+\cos ^{2} \theta}} \text { and } p_{2}=\frac{|-5|}{\sqrt{\operatorname{cosec}^{2} \theta+\sec ^{2} \theta}} \\
\therefore \mathrm{p}_{1}^{2}=\frac{25 \cos ^{2} 2 \theta}{1} \text { and } 4 \mathrm{p}_{2}^{2} &=\frac{4(25)}{\operatorname{cosec}^{2} \theta+\sec ^{2} \theta} \\
\therefore \mathrm{p}_{1}^{2}+4 \mathrm{p}_{2}^{2} &=25\left(\cos ^{2} \theta-\sin ^{2} \theta\right)^{2}+\frac{100}{\left(\frac{1}{\sin ^{2} \theta}\right)+\frac{1}{\cos ^{2} \theta}} \\
&=25(\cos \theta-\sin \theta)^{2}(\cos \theta+\sin \theta)^{2}+100 \sin ^{2} \theta \cos ^{2} \theta \\
&=25(1-\sin 2 \theta)(1+\sin 2 \theta)+25\left(4 \sin ^{2} \theta \cos ^{2} \theta\right) \\
&=25\left(1-\sin 2 \theta+\sin 2 \theta-\sin ^{2} 2 \theta\right)+25(2 \sin \theta \cos \theta)^{2} \\
&=25\left(1-\sin ^{2} 2 \theta\right)+25(\sin 2 \theta)^{2} \\
&=25\left(\cos ^{2} 2 \theta\right)+25\left(\sin ^{2} 2 \theta\right)=25
\end{aligned}
$$
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