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  • If p and q are the lengths of perpendiculars from the origin to the lines x cos θ - y sin θ=k cos 2θ and x sec θ + y cosec θ = k, respectively, prove that p^2 + 4q^2 = k^2.

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. 

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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

Posted by

Gautam harsolia

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