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Q.

If e| and e are the eccentricities of the hyperbola x2a2y2b2=1  and its conjugate hyperbola, the value of 1e2+1e|2 is

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a

2

b

1

c

0

d

3

answer is C.

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Detailed Solution

e=a2+b2a2then e|=a2+b2b2

1e2+1(e|)2=a2+b2a2+b2=1

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