∵∠ FBF ′=90∘∴(a2e2+b2)2+(a2e2+b2)2=(2ae)2⇒2(a2e2+b2)=4a2e2⇒e2=b2/a2 Also e2=1−b2/a2=1−e2⇒2e2=1,e=21.
An ellipse has OB as semi minor axis, F and F′ its focii and the angle FBF' is a right angle. Then the eccentricity of the ellipse is
Held on 30 Apr 2005 · Verified 6 Jul 2026.
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