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  4. The angle between the pair of lines (x2+y2) sin 2α =(x cos θ -y sin θ )2 is :

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Q. The angle between the pair of lines $ ({{x}^{2}}+{{y}^{2}}){{\sin }^{2}}\alpha ={{(x\,\cos \theta -y\,\sin \theta )}^{2}} $ is :

J & K CETJ & K CET 2007Straight Lines

Solution:

Given pair of lines of $ ({{x}^{2}}+{{y}^{2}})si{{n}^{2}}\alpha ={{(x\,\cos \,\theta -y\,\sin \theta )}^{2}} $
$ \Rightarrow $ $ {{x}^{2}}\,{{\sin }^{2}}\alpha +{{y}^{2}}\,{{\sin }^{2}}\alpha ={{x}^{2}}{{\cos }^{2}}\theta $
$ +{{y}^{2}}\,{{\sin }^{2}}\theta -2xy\,\sin \theta \,\cos \theta $
$ \Rightarrow $ $ {{x}^{2}}({{\sin }^{2}}\alpha -{{\cos }^{2}}\theta )+{{y}^{2}}({{\sin }^{2}}\alpha -{{\sin }^{2}}\theta ) $
$ +2xy\,\,\sin \theta \,\cos \theta =0 $
$ \Rightarrow $ $ {{x}^{2}}({{\sin }^{2}}\alpha -{{\cos }^{2}}\theta )+{{y}^{2}}({{\sin }^{2}}\alpha -{{\sin }^{2}}\theta ) $
$ +2(\sin \theta \,\cos \theta )\,xy=0 $
On comparing with $ a{{x}^{2}}+b{{y}^{2}}+2hxy=0, $
We get, $ a={{\sin }^{2}}\alpha -{{\cos }^{2}}\theta ,\,b={{\sin }^{2}}\alpha -{{\sin }^{2}}\theta $
and $ h=\sin \,\theta \,\cos \theta $
Let $ \theta $ be the angle between the pair of lines. $ \therefore $
$ \tan \theta =\left| \frac{2\,\sqrt{{{h}^{2}}-ab}}{a+b} \right| $
$ =\left| \frac{2\sqrt{\begin{align} & {{\sin }^{2}}\,\theta \,{{\cos }^{2}}\theta -({{\sin }^{2}}\alpha -{{\cos }^{2}}\theta ) \\ & \times ({{\sin }^{2}}\alpha -{{\sin }^{2}}\theta ) \\ \end{align}}}{{{\sin }^{2}}\alpha -{{\cos }^{2}}\theta +{{\sin }^{2}}\alpha -{{\sin }^{2}}\theta )} \right| $
$ =\left| \frac{2\sqrt{\begin{align} & {{\sin }^{2}}\theta \,{{\cos }^{2}}\theta -{{({{\sin }^{2}}\alpha )}^{2}}+{{\sin }^{2}}\alpha \\ & {{\sin }^{2}}\theta +{{\sin }^{2}}\alpha \,{{\cos }^{2}}\theta -{{\sin }^{2}}\theta \,{{\cos }^{2}}\theta \\ \end{align}}}{-(-1-2{{\sin }^{2}}\alpha )} \right| $
$ =\left| \frac{2\sqrt{{{\sin }^{2}}\alpha ({{\sin }^{2}}\theta +{{\cos }^{2}}\theta )-{{(si{{n}^{2}}\alpha )}^{2}}}}{-\cos \,2\,\,\alpha } \right| $
$ =\left| \frac{2\sqrt{{{\sin }^{2}}\alpha (1-{{\sin }^{2}}\alpha )}}{-\cos \,2\alpha } \right| $
$ \Rightarrow $ $ \tan \theta =\left| \frac{\sin \,2\alpha }{\cos \,2\alpha } \right|=\tan \,2\alpha $
$ \Rightarrow $ $ \theta =2\alpha $