A) \[\theta \]
B) \[2\,\theta \]
C) \[\alpha \]
D) \[2\,\alpha \]
Correct Answer: D
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. \[\therefore \] \[\tan \theta =\left| \frac{2\,\sqrt{{{h}^{2}}-ab}}{a+b} \right|\] \[=\left| \frac{2\sqrt{\begin{align} & {{\sin }^{2}}\,\theta \,{{\cos }^{2}}\theta -({{\sin }^{2}}\alpha -{{\cos }^{2}}\theta ) \\ & \times ({{\sin }^{2}}\alpha -{{\sin }^{2}}\theta ) \\ \end{align}}}{{{\sin }^{2}}\alpha -{{\cos }^{2}}\theta +{{\sin }^{2}}\alpha -{{\sin }^{2}}\theta )} \right|\] \[=\left| \frac{2\sqrt{\begin{align} & {{\sin }^{2}}\theta \,{{\cos }^{2}}\theta -{{({{\sin }^{2}}\alpha )}^{2}}+{{\sin }^{2}}\alpha \\ & {{\sin }^{2}}\theta +{{\sin }^{2}}\alpha \,{{\cos }^{2}}\theta -{{\sin }^{2}}\theta \,{{\cos }^{2}}\theta \\ \end{align}}}{-(-1-2{{\sin }^{2}}\alpha )} \right|\] \[=\left| \frac{2\sqrt{{{\sin }^{2}}\alpha ({{\sin }^{2}}\theta +{{\cos }^{2}}\theta )-{{(si{{n}^{2}}\alpha )}^{2}}}}{-\cos \,2\,\,\alpha } \right|\] \[=\left| \frac{2\sqrt{{{\sin }^{2}}\alpha (1-{{\sin }^{2}}\alpha )}}{-\cos \,2\alpha } \right|\] \[\Rightarrow \] \[\tan \theta =\left| \frac{\sin \,2\alpha }{\cos \,2\alpha } \right|=\tan \,2\alpha \] \[\Rightarrow \] \[\theta =2\alpha \]
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