A) \[\frac{4}{3}\]
B) 4
C) \[2\]
D) \[\frac{2}{\sqrt{3}}\]
Correct Answer: C
Solution :
We know that, eccentricity of the hyperbola \[\frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{{{b}^{2}}}=1\] is \[e'=\sqrt{\frac{{{a}^{2}}+{{b}^{2}}}{{{b}^{2}}}}\] Given, equation of hyperbola is \[{{x}^{2}}-3{{y}^{2}}=1\] \[\Rightarrow \] \[\frac{{{x}^{2}}}{1}-\frac{{{y}^{2}}}{1/3}=1\] Here, \[{{a}^{2}}=1\] and \[{{b}^{2}}=1\] \[\therefore \] \[e'=\sqrt{\frac{1+\frac{1}{3}}{\frac{1}{3}}}\] \[=\sqrt{\frac{3+1}{1}}\] \[=\sqrt{4}=2\]
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