Question

If $p,q$ are the perpendicular distances from the origin to the lines $x\sec \alpha +y\text{cosec}\alpha =a$ and $x\cos \alpha -y\sin \alpha =a\cos 2\alpha ,$ then $4p^2+q^2=$

• A. $a^2$
• B. $4a^2$
• C. $2a^2$
• D. $6a^2$

Answer 1

The correct option is A $a^2$

The perpendicular distance from origin to the line $ax+by+c=0$ is given by $\frac{|c|}{\sqrt{a^2+b^2}}$.

Distance from origin to line $x\sec \alpha +y\text{cosec}\alpha =a$ is $p.$

So, $p=\left|\frac{a}{\sqrt{{\sec }^2\alpha +{\text{cosec}}^2\alpha }}\right|=\left|\frac{2a\cos \alpha \sin \alpha }{2}\right|=\left|\frac{a}{2}\sin 2\alpha \right|\to \left(1\right)$

& similarly, $q=\left|\frac{a\cos 2\alpha }{\sqrt{{\cos }^2\alpha +{\sin }^2\alpha }}\right|=|a\cos 2\alpha |\to \left(2\right)$

Square and add $\left(1\right)$ & $\left(2\right)$, we get

$\Rightarrow (2p{)}^2+q^2=a^2$

$\Rightarrow 4p^2+q^2=a^2$