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  • The straight lines represented by the equation <img alt="\mathrm{\left(x^2+y^2\right) \sin ^2 \alpha=(x \cos \theta-y \sin \theta)^2}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm

The straight lines represented by the equation \mathrm{\left(x^2+y^2\right) \sin ^2 \alpha=(x \cos \theta-y \sin \theta)^2} are inclined to each other at an angle equal to
 

Option: 1

2 \alpha


 


Option: 2

\pi / 2+\alpha
 


Option: 3

\alpha
 


Option: 4

\alpha / 2


Answers (1)

best_answer

The given equation is \mathrm{\left(\cos ^2 \theta-\sin ^2 \alpha\right) \mathrm{x}^2+\left(\sin ^2 \theta-\sin ^2 \alpha\right) \mathrm{y}^2-2 \sin \theta \cos \theta \mathrm{xy}=0 }

\mathrm{\therefore } Angle between the lines

\mathrm{ =\tan ^{-1}\left|\frac{2 \sqrt{h^2-a b}}{a+b}\right|=\tan ^{-1}\left|\frac{2 \sqrt{\sin ^2 \theta \cos ^2 \theta-\left(\cos ^2 \theta-\sin ^2 \alpha\right)\left(\sin ^2 \theta-\sin ^2 \alpha\right)}}{\left(\cos ^2 \theta-\sin ^2 \alpha\right)+\left(\sin ^2 \theta-\sin ^2 \alpha\right)}\right| }

\mathrm{ =\tan ^{-1} \frac{2 \sin \alpha \cos \alpha}{1-2 \sin ^2 \alpha}=2 \alpha }

Hence option 1 is correct.



 

Posted by

sudhir.kumar

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