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Question

If e1 and e2 are the eccentricities of a hyperbola3x2-3y2=25 and its conjugate, then


A

e12+e22=2

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B

e12+e22=4

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C

e1+e2=4

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D

e1+e2=2

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Solution

The correct option is B

e12+e22=4


Finding the value of the given expression:

Given equation can be rewritten as x2y2=253

Here, a2=1,b2=1

e1=1+b2a2=1+1=2

The equation of the conjugate hyperbola is x2+y2=253

e2=1+a2b2=1+1=2e12+e22=(2)2+(2)2=2+2=4

Hence, the correct answer is option (B).


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