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Home  >>  CBSE XI  >>  Math  >>  Straight Lines
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If $p$ and $q$ are the lengths of perpendiculars from the origin to the lines $ x \cos \theta - y \sin \theta = k \cos 2\theta$ and $ x \sec \theta + y \: cosec \theta -k$, respectively, prove that $ p^2+4q^2=k^2$

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  • The perpendicular distance (d) of the line $Ax+By+C=0$ from the point $(x_1, y_1)$ is $ d = \large\frac{|Ax_1+By_1+C|}{\sqrt {A^2+B^2}}$
The equation of the given lines are
$ x \cos \theta - y \sin \theta = k \cos 2 \theta $------------(1)
$x \sec \theta +y \: cosec \theta = k$-------------------(2)
Comparing eqn (1) to the general equation of the line $Ax+By+C=0$ we get,
$A = \cos \theta, B = -\sin \theta \: and \: C = -k \cos 2 \theta$
If $p$ is the perpendicular distance from the origin to line (1) then
$ p = \large\frac{|A(0)+B(0)+C|}{\sqrt {A^2+B^2}}$$ = \large\frac{|C|}{\sqrt{A^2+B^2}}$
Substituting the values,
$ p = \large\frac{-k \cos 2 \theta}{\sqrt{\sin^2 \theta+ \cos^2\theta}}$$ = |-k \cos 2 \theta|$-----------(1)
Comparing eqn (2) to the general equation of the line $Ax+By+C=0$ we get,
$ A = \sec \theta , B = cosec \theta$, and $ C = -k$
$q$ is the length of the perpendicular from the origin (0,0)
$ \therefore q = \bigg| \large\frac{A(0)+B(0)+C}{\sqrt{A^2+B^2}} \bigg|$$ = \large\frac{|C|}{\sqrt{A^2+B^2}}$
$ = \large\frac{|-k|}{\sqrt{\sec^2 \theta+ cosec^2 \theta}}$------------(2)
$ \therefore $ from eqn (1) and (2) we have,
$p^2+4q^2=(1-k \cos 2 \theta|^2+4 \bigg( \large\frac{|-k|}{\sqrt{\sec^2\theta+ cosec^2\theta}} \bigg)^2$
$ = k^2\cos^22 \theta+ \large\frac{4k^2}{\Large\frac{1}{\cos^2\theta}+\Large\frac{1}{\sin^2\theta}}$
$ = k^2\cos^22\theta + \large\frac{4k^2}{\Large\frac{\sin^2\theta+ \cos^2\theta}{\sin^2\theta\cos^2\theta}}$
But $ \sin^2\theta + \cos^2 \theta=1$
$ = k^2\cos^22\theta+ \large\frac{4k^2}{\Large\frac{1}{\sin^2\theta \cos^2\theta}}$
$ = k^2 \cos^22\theta + 4 k^2 \sin^2\theta \cos^2\theta$
$ = k^2\cos^2 2 \theta + k^2(2 \sin \theta \cos \theta )^2$
But $2 \sin \theta \cos \theta = \sin^2\theta$
$ = k^2\cos^22\theta+k^2\sin^22\theta$
$ = k^2 ( \sin^22\theta + \cos^2 2 \theta)$
$ = k^2$
Hence $p^2+4q^2=k^2$ is proved.
answered May 13, 2014 by thanvigandhi_1
 

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