A) \[\frac{2}{\sqrt{3}}\]
B) \[\sqrt{3}\]
C) 2
D) None of these
Correct Answer: C
Solution :
Given, equation of hyperbola is \[{{x}^{2}}-3{{y}^{2}}=2x+8\] Þ \[{{x}^{2}}-2x-3{{y}^{2}}=8\] Þ \[{{(x-1)}^{2}}-3{{y}^{2}}=9\] Þ \[\frac{{{(x-1)}^{2}}}{9}-\frac{{{y}^{2}}}{3}=1\] Conjugate of this hyperbola is \[-\frac{{{(x-1)}^{2}}}{9}+\frac{{{y}^{2}}}{3}=1\] and its eccentricity \[(e)=\sqrt{\left( \frac{{{a}^{2}}+{{b}^{2}}}{{{b}^{2}}} \right)}\] Here, \[{{a}^{2}}=9\], \[{{b}^{2}}=3\]; \[\therefore \] \[e=\sqrt{\frac{9+3}{3}}=2\].
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