A) \[\sqrt{3}\]
B) \[\sqrt{2}\]
C) \[1/\sqrt{2}\]
D) 2
Correct Answer: B
Solution :
Hyperbola is \[\frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{{{b}^{2}}}=1\]. Here, transverse and conjugate axis of a hyperbola is equal. i.e., \[a=b\]\[\therefore {{x}^{2}}-{{y}^{2}}={{a}^{2}}\]; which is a rectangular hyperbola. Hence, eccentricity \[e=\sqrt{1+\frac{{{b}^{2}}}{{{a}^{2}}}}=\sqrt{2}\].
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