A) 1
B) 2
C) 3
D) 4
Correct Answer: B
Solution :
[b] Conjugate hyperbola of hyperbola \[{{x}^{2}}-3{{y}^{2}}=1\text{ }be\text{ }3{{x}^{2}}-{{y}^{2}}=1\] Now, \[e=a.\sqrt{1+\frac{{{b}^{2}}}{{{a}^{2}}}}=\sqrt{3+1}=\sqrt{4}=2\].You need to login to perform this action.
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