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Question

The eccentricity of the hyperbola conjugate to x23y2=2x+8 is

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Solution

The given equation reduces to (x1)29y23=1. Its conjugate hyperbola is y23(x1)29=1; thus a2=9,b2=3
Using a2=b2(e21), we get
9=3(e21)e=2.

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