Published by:
CGP EDU Academic Team
Published on: August 13, 2026
If AB is a double ordinate of the hyperbola 
= 1 such that Δ OAB (O is the origin) is an equilateral triangle, then the eccentricity ‘e’ of the hyperbola satisfies-
A
e> 
B
1 < e < 
C
e = 
D
e > 
Text Solution
Verified by ExpertsThe correct answer is:
D
Let AB = 2 l , then AM = 1
Thus, we have A = 
Now, since OAB is an equilateral triangle,
Therefore we have, OA = 2 l i.e. OM 2 + AM 2
= (2 l ) 2 i.e. 
+ l 2 = 4 l 2 Gives l
2 = 
> 0 i.e. 3b 2 – a 2 > 0 i.e. 3a 2 (e 2 – 1) > a 2 Gives e > 
.
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