In standard ellipse a2x2+b2y2=1
Semi-major axis OA=a
Semi-minor axis OB=b
Focus F1≡(−ae,0)
F2≡(ae,0)

∠F1BF2=90∘
OF1=OF2⇒∠F1BO=∠F2BO=45∘
⇒bae=tan45∘=1
⇒b=ae
Also, b2=a2(1−e2)
⇒a2e2=a2(1−e2)
⇒2e2=1
⇒e2=21
JEE Main 2014 — Mathematics Coordinate Geometry
If OB is the semi-minor axis of an ellipse, F1 and F2 are its focii and the angle between F1B and F2B is a right angle, then the square of the eccentricity of the ellipse is
Held on 9 Apr 2014 · Verified 6 Jul 2026.
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