Question

If $p$ and $p^{\prime }$ be the lengths of perpendiculars from origin
to the lines $x\sec \theta -y\cos \theta =a$ and $x\cos \theta -y\sin \theta =a\cos 2\theta$ respectively, then prove
that $4p^2+p^{\prime 2}=a^a$.

Answer 1

$P=\frac{-a}{\surd ({\sec }^2\theta +{\csc }^2\theta )}$
$\therefore 4p^2=\frac{4a^2{\sin }^2\theta {\cos }^2\theta }{{\sin }^2\theta +{\cos }^2\theta }$
Or $4p^2=a^2{\left(2\sin \theta \cos \theta \right)}^2{a}^2{\sin }^{2}2\theta$
$P^{\prime }=\frac{-a\cos 2\theta }{\surd ({\cos }^{2}2\theta +{\sin }^2\theta )}$
$\therefore {p^{\prime }}^2=a^2{\cos }^{2}2\theta$
$\therefore 4p^2+{p^{\prime }}^2=a^2({\sin }^{2}2\theta +{\cos }^{2}2\theta )=a^2$