Question

If $p_1$ and $p_2$ are the lengths of the perpendiculars from the origin to the straight lines $x\sec \theta +y\text{cosec}\theta =a$ and $x\cos \theta -y\sin \theta =a\cos 2\theta$ respectively, then the value of $4p_1^2+p_2^2$ is

• A. $4a^2$
• B. $2a^2$
• C. $a^2$
• D. $\frac{a^2}{2}$

Answer 1

The correct option is C $a^2$

Given line: $x\sec \theta +y\text{cosec}\theta =a$
$\Rightarrow x\sin \theta +y\cos \theta =a\sin \theta \cos \theta \Rightarrow x\cos (\pi /2-\theta )+y\sin (\pi /2-\theta )=a\sin \theta \cos \theta$
$p_1=|a\sin \theta \cos \theta |\left(\text{normal form}\right)\Rightarrow p_1^2=a^2{\sin }^2\theta {\cos }^2\theta \Rightarrow 4p_1^2=a^2{\sin }^{2}2\theta ...\left(i\right)$
And,
$x\cos \theta -y\sin \theta =a\cos 2\theta \Rightarrow x\cos (-\theta )+y\sin (-\theta )=a\cos 2\theta$
, $p_2=|a\cos 2\theta |$
$\Rightarrow p_2^2=a^2{\cos }^{2}2\theta ...\left(ii\right)$

On adding Eqs. $\left(i\right)$ and $\left(ii\right)$, we get
$4p_1^2+p_2^2=a^2$