Question

If p and q are the length of perpendiculars from the origin to the lines $xcos\theta -ysin\theta =kcos2\theta$ and $xsec\theta +ycosec\theta =k$ respectively, prove that $p^2+4q^2=k^2$

Answer 1

Length of perpendicular from origin to line $xcos\theta -ysin\theta -kcos2\theta =0$ is

$p=\left|\frac{0\times cos\theta -0\times sin\theta -kcos2\theta }{\sqrt{co{s}^2\theta +si{n}^2\theta }}\right|$

$=\left|\frac{-kcos2\theta }{1}\right|=kcos2\theta$

Length of perpendicular from origin to line x $sec\theta +ycosec\theta -k=0$ is

$q=\left|\frac{0\times sec\theta +0\times cosec\theta -k}{\sqrt{se{c}^2\theta +cose{c}^2\theta }}\right|$

$=\left|\frac{-k}{\sqrt{\frac{si{n}^2\theta +co{s}^2\theta }{si{n}^2\theta co{s}^2\theta }}}\right|$

$=|-ksin\theta cos\theta |=\frac{k}{2}sin2\theta$

Now

$p^2+4q^2=(kcos2\theta {)}^2+4{\left(\frac{k}{2}sin2\theta \right)}^2$

$=kco{s}^{2}2\theta +\frac{4}{4}{k}^{2}si{n}^{2}2\theta$

$=k^2(co{s}^{2}2\theta +si{n}^{2}2\theta )=k^2$