A hyperbola has its centre at the origin, passes through the point (4,2) and has transverse axis of length 4 along the x-axis. Then the eccentricity of the hyperbola is :
A
√3
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B
2√3
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C
2
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D
32
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Solution
The correct option is B2√3 Given 2a=4 ⇒a=2
The hyperbola x2a2−y2b2=1 passes through (4,2)
So, 4222−22b2=1 ⇒b2=43 ∴c2=a2+b2=22+43 ⇒c2=163 ⇒e=ca=√1632=2√3