Q.

If e1 and e2 are the eccentricities of x2a2+y2b2=1 and x2b2+y2a2=1respectively  then

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a

1e12+1e22=1

b

e1=e2

c

1e1+1e2=1

d

e1e2=1

answer is A.

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Detailed Solution

Let a>b Now eccentricities of given hyperbolas are e1=a2-b2a2 e2=a2-b2a2 e1=e2

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