A) \[\frac{1}{e}\]
B) \[\frac{-1}{e}\]
C) \[e\]
D) \[\frac{2}{e}\]
Correct Answer: A
Solution :
The equations of the asymptotes of the hyperbola \[\frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{{{b}^{2}}}=1\]are\[y=\pm \frac{b}{a}ie\], \[y=\frac{b}{a}\]and\[y=\frac{-b}{a}x\] If\[\theta \]is the angle between the two asymptotes, then,\[\tan \theta =\frac{\frac{b}{a}+\frac{b}{a}}{1-\frac{b}{a}\cdot \frac{b}{a}}=\frac{2b}{a}\times \frac{{{a}^{2}}}{{{a}^{2}}-{{b}^{2}}}=\frac{2ab}{{{a}^{2}}-{{b}^{2}}}\] From this, we get\[\cos \theta =\frac{{{a}^{2}}-{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}}\] \[\Rightarrow \] \[2{{\cos }^{2}}\frac{\theta }{2}-1=\frac{{{a}^{2}}-{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}}\] \[\Rightarrow \] \[2{{\cos }^{2}}\frac{\theta }{2}=1+\frac{{{a}^{2}}-{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}}\] \[\Rightarrow \] \[2{{\cos }^{2}}\frac{\theta }{2}=\frac{2{{a}^{2}}}{{{a}^{2}}+{{b}^{2}}}\] \[\Rightarrow \] \[{{\cos }^{2}}\frac{\theta }{2}=\frac{{{a}^{2}}}{{{a}^{2}}+{{b}^{2}}}\] But\[{{b}^{2}}={{a}^{2}}({{e}^{2}}-1)={{a}^{2}}{{e}^{2}}-{{a}^{2}}\Rightarrow {{a}^{2}}+{{b}^{2}}={{a}^{2}}{{e}^{2}}\] \[\therefore \] \[{{\cos }^{2}}\left( \frac{\theta }{2} \right)=\frac{{{a}^{2}}}{{{a}^{2}}{{e}^{2}}}=\frac{1}{{{e}^{2}}}\] \[\Rightarrow \] \[\cos \left( \frac{\theta }{2} \right)=\frac{1}{e}\]
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