If $p$ and $q$ are lengths of the perpendiculars from 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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Aicharya · Accepted Answer

$p=\frac{|0+0-k|}{\sqrt{\sec ^2 \theta+\operatorname{cosec}^2 \theta}}, q=\frac{\left|0-0-k \cos ^2 \theta\right|}{\sqrt{\cos ^2 \theta+\sin ^2 \theta}}$ $\Rightarrow \quad p^2=\frac{k^2}{\sec ^2 \theta+\operatorname{cosec}^2 \theta}, q^2=k^2 \cos ^2 2 \theta$ $\Rightarrow \quad p^2=\frac{k^2}{\frac{1}{\cos ^2 \theta}+\frac{1}{\sin ^2 \theta}}, q^2=k^2 \cos ^2 2 \theta$ $\Rightarrow \quad p^2=k^2 \sin ^2 \theta \cdot \cos ^2 \theta, q^2=k^2 \cos ^2 2 \theta$ $\Rightarrow \quad p^2=\frac{k^2}{4} \sin ^2 2 \theta, q^2=k^2 \cos ^2 2 \theta$ So, $p^2+\frac{q^2}{4}=\frac{k^2}{4} \Rightarrow 4 p^2+q^2=k^2$

If \(p\) and \(q\) are lengths of the perpendiculars from origin to the lines \(x \sec (\theta)+y…

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