Question

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

• A. $1$
• B. $a^2$
• C. $\frac{1}{a^2}$
• D. $a$
• E. $\frac{1}{a}$

Answer 1

The correct option is B $a^2$

We have,
$p_1=$ Length of the perpendicular from $(0,0)$ to $x\sec \theta +y\csc \theta =a$
$\Rightarrow p_1=\left|\frac{0\sec \theta +0\csc \theta -a}{\sqrt{{\sec }^2\theta +{\csc }^2\theta }}\right|$
$=\frac{a\cos \theta \sin \theta }{\sqrt{{\sin }^2\theta +{\cos }^2\theta }}$
$=a\sin \theta \cos \theta$
$\Rightarrow 2p_1=2a\sin \theta \cos \theta$
$\Rightarrow 2p_1=a\sin 2\theta$
and $p_2=$ Length of the perpendicular from $(0,0)$ to
$x\cos \theta -y\sin \theta -a\cos 2\theta =0$
$\Rightarrow p_2=\left|\frac{0\cos \theta -0\sin \theta -a\cos 2\theta }{\sqrt{{\cos }^2\theta +{\sin }^2\theta }}\right|$
$\Rightarrow p_2=a\cos 2\theta$
Therefore, $4p_1^2+p_2^2={\left(a\sin 2\theta \right)}^2+{\left(a\cos 2\theta \right)}^2=a^2$