A) \[{{\cos }^{-1}}\left( \frac{4}{5} \right)\]
B) \[{{\sin }^{-1}}\left( \frac{4}{5} \right)\]
C) \[{{\tan }^{-1}}\left( \frac{4}{5} \right)\]
D) None of these
Correct Answer: B
Solution :
The equation of pair of tangent from the point \[\left( 1,\,\,\frac{1}{2} \right)\]to the circle \[S\equiv {{x}^{2}}+{{y}^{2}}+4x+2y-4=0\]is \[S{{S}_{1}}={{T}^{2}}\] \[({{x}^{2}}+{{y}^{2}}+4x+2y-4)\left( 1+\frac{1}{4}+4+1-4 \right)\] \[={{\left[ x+\frac{1}{2}y+2(x+1)+y+\frac{1}{2}-4 \right]}^{2}}\] \[\frac{9}{4}({{x}^{2}}+{{y}^{2}}+4x+2y-4)={{\left( 3x+\frac{3}{2}y-\frac{3}{2} \right)}^{2}}\] \[\Rightarrow \]\[\frac{9}{4}({{x}^{2}}+{{y}^{2}}+4x+2y-4)=\frac{9}{4}{{(2x+y-1)}^{2}}\] \[\Rightarrow \] \[{{x}^{2}}+{{y}^{2}}+4x+2y-4\] \[=4{{x}^{2}}+{{y}^{2}}+1+4xy-2y-4x\] \[\Rightarrow \] \[3{{x}^{2}}+4xy-8x-4y+5=0\] On comparing this equation with \[a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0\], we get \[a=3,\,\,h=2,\,\,b=0,\,\,g=-4,\,\,f=-2,\,\,c=5\] Required angle\[={{\tan }^{-1}}\left| \frac{2\sqrt{{{h}^{2}}-ab}}{a+b} \right|\] \[={{\tan }^{-1}}\left| \frac{2\sqrt{4-0}}{3} \right|\] \[={{\tan }^{-1}}\frac{4}{3}\] \[={{\sin }^{-1}}\left( \frac{4}{5} \right)\]
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