CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

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 $$4p^{ 2 }+p' ^{ 2 }=a^{ a }$$.


Solution

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

Mathematics

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image