# Q : 16     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  $x\sec \theta +y\hspace{1mm}cosec\hspace{1mm}\theta =k$ , respectively,  prove that  $p^2+4q^2=k^2$.

G Gautam harsolia

Given equations of lines are    $x\cos \theta -y\sin \theta =k\cos 2\theta$  and  $x\sec \theta +y\hspace{1mm}cosec\hspace{1mm}\theta =k$

We can rewrite the equation $x\sec \theta +y\hspace{1mm}cosec\hspace{1mm}\theta =k$ as

$x\sin \theta +y\cos \theta = k\sin\theta\cos\theta$
Now, we know that

$d = \left | \frac{Ax_1+By_1+C}{\sqrt{A^2+B^2}} \right |$

In equation $x\cos \theta -y\sin \theta =k\cos 2\theta$

$A= \cos \theta , B = -\sin \theta , C = - k\cos2\theta \ and \ (x_1,y_1)= (0,0)$

$p= \left | \frac{\cos\theta .0-\sin\theta.0-k\cos2\theta }{\sqrt{\cos^2\theta+(-\sin\theta)^2}} \right | = |-k\cos2\theta|$
Similarly,
in the equation $x\sin \theta +y\cos \theta = k\sin\theta\cos\theta$

$A= \sin \theta , B = \cos \theta , C = -k\sin\theta\cos\theta \ and \ (x_1,y_1)= (0,0)$

$q= \left | \frac{\sin\theta .0+\cos\theta.0-k\sin\theta\cos\theta }{\sqrt{\sin^2\theta+\cos^2\theta}} \right | = |-k\sin\theta\cos\theta|= \left | -\frac{k\sin2\theta}{2} \right |$
Now,

$p^2+4q^2=(|-k\cos2\theta|)^2+4.(|-\frac{k\sin2\theta}{2})^2= k^2\cos^22\theta+4.\frac{k^2\sin^22\theta}{4}$
$=k^2(\cos^22\theta+\sin^22\theta)$
$=k^2$
Hence proved

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