Find the eccentricity of the hyperbola x2a2−y2b2=1 which passes through (4,0) & (3√2,2)
A
√2
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B
√3
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C
√5
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D
−√3
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Solution
The correct option is B√3 Since, given hyperbola passes through (4,0) & (3√2,2) ⇒16a2−0b2=1⇒a2=16 And 1816−4b2=1⇒b2=32 Hence, required eccentricity is =√1+b2a2=√3