A) \[{{\tan }^{-1}}\left( -\frac{1}{2} \right)\]
B) \[{{\tan }^{-1}}2\]
C) \[{{\tan }^{-1}}\frac{1}{2}\]
D) \[{{60}^{o}}\]
Correct Answer: B
Solution :
From \[x+y=1,\]to make the curve \[{{x}^{2}}+{{y}^{2}}-2x-1=0\] homogenous. \[\Rightarrow {{x}^{2}}+{{y}^{2}}-2x(x+y)-{{(x+y)}^{2}}=0\] \[\therefore 2{{x}^{2}}+4xy=0\] or \[{{x}^{2}}+2xy=0\] \[\therefore \tan \theta =\frac{2\sqrt{{{h}^{2}}-ab}}{a+b}\] and \[a=1,\ b=0,\ h=1\] \[\therefore \tan \theta =\frac{2\sqrt{{{1}^{2}}-0}}{1}\Rightarrow \theta ={{\tan }^{-1}}(2)\].
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