If $$p$$ and $$p'$$ 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 $$4p^{ 2 }+p' ^{ 2 }=a^{ a }$$.
$$P=\dfrac { -a }{ \surd \left( \sec ^{ 2 }{ \theta } +\csc ^{ 2 }{ \theta } \right) }$$
$$\therefore\quad 4{ p }^{ 2 }=\dfrac { 4a^{ 2 }\sin ^{ 2 }{ \theta } \cos ^{ 2 }{ \theta } }{ \sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } }$$
Or $$4{ p }^{ 2 }={ a }^{ 2 }{ \left( 2\sin { \theta } \cos { \theta } \right) }^{ 2 }{ a }^{ 2 }\sin ^{ 2 }{ 2\theta }$$
$$P'=\dfrac { -a\cos { 2\theta } }{ \surd \left( \cos ^{ 2 }{ 2\theta } +\sin ^{ 2 }{ \theta } \right) }$$
$$\therefore\quad { p' }^{ 2 }={ a }^{ 2 }\cos ^{ 2 }{ 2\theta }$$
$$\therefore\quad 4{ p }^{ 2 }+{ p' }^{ 2 }={ a }^{ 2 }\left( \sin ^{ 2 }{ 2\theta } +\cos ^{ 2 }{ 2\theta } \right) ={ a }^{ 2 }$$
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