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27. For eccentricity in hyperbola (e) which relation is correct?(1 Point)e < 1e = 1e > 1e = 0​

Answer»

Equation of HYPERBOLA is a 2 x 2 − b 2 y 2 =1∵b 2 =a 2 (E 2 −1)⇒e 2 =1+ a 2 b 2 = a 2 a 2 +b 2 Equation of conjugate hyperbola is a 2 x 2 − b 2 y 2 =−1⇒ b 2 y 2 − a 2 x 2 =1∴e ′ 2 =1+ b 2 a 2 = b 2 a 2 +b 2 Now, e 2 1 + e ′ 2 1 = a 2 +b 2 a 2 +b 2 =1


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