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Question: Answered & Verified by Expert
If e1 and e2 are the eccentricities of a hyperbola 3x2-3y2=25 and its conjugate, then
MathematicsHyperbolaJEE Main
Options:
  • A e12+e22=2
  • B e12+e22=4
  • C e1+e2=4
  • D e1+e2=2
Solution:
2054 Upvotes Verified Answer
The correct answer is: e12+e22=4

Given equation can be written as

x2-y2=253

 e1=1+b2a2

=1+1=2

The equation of conjugate hyperbola is

-x2+y2=253

 e2=1+b2a2=1+1=2

 e12+e22=22+22=4

Hence, e12+e22=4.

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