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Question

If e and e1 are the eccentricities of a hyperbola and its conjugate, then 1e2+1e21 is equal to

A
1
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B
0
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C
1
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D
None of these
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Solution

The correct option is C 1
Consider Hyperbola x2a2y2b2=1..(1)
So it's conjugate hyperbola is y2b2x2a2=1..(2)
Given e and e1 are eccentricties of (1) and (2) respectively
e2=1+b2a2=a2+b2a2
& e21=1+a2b2=a2+b2b2
Thus 1e2+1e21=a2a2+b2+b2a2+b2=1
Hence, option 'C' is correct.

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