For any standard hyperbola we know that
eccentricity, e=1+a2b2, conjugate axis =2b and distance between the foci =2ae.
Given, 2b=21(2ae).
⇒ ab=2e
Now, e=1+a2b2=1+4e2
⇒e2=1+4e2
⇒43e2=1
⇒e2=34
⇒e=32
JEE Main 2016 — Mathematics Coordinate Geometry
The eccentricity of the hyperbola whose length of its conjugate axis is equal to half of the distance between its foci, is
Held on 3 Apr 2016 · Verified 6 Jul 2026.
32
3
34
34
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