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Question

Find the eccentricity of the hyperbola x2a2y2b2=1 which passes through (4,0) & (32,2)

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) & (32,2)
16a20b2=1a2=16
And 18164b2=1b2=32
Hence, required eccentricity is =1+b2a2=3

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