A) 2
B) \[\frac{2}{\sqrt{3}}\]
C) 4
D) \[\frac{4}{3}\]
Correct Answer: A
Solution :
Eccentricity of \[\frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{{{b}^{2}}}=1\]is \[e=\sqrt{\frac{{{a}^{2}}+{{b}^{2}}}{{{a}^{2}}}}\] Eccentricity of conjugate hyperbola, \[e'=\sqrt{\frac{{{a}^{2}}+{{b}^{2}}}{{{b}^{2}}}}\] Write the given equation in standard form, \[\frac{{{x}^{2}}}{1}-\frac{{{y}^{2}}}{1/3}=1\]Þ\[{{a}^{2}}=1,\,\,{{b}^{2}}=\frac{1}{3}\] \\[e'=\sqrt{\frac{1+1/3}{1/3}}=\sqrt{4}=2\].
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