Let S and S′ be the two foci of the ellipse x2a2+y2b2=1. If the circle described on SS′ as diameter touches the ellipse in real points, then 6e2=(where e is eccentricity of ellipse)
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Solution
∵ Circle with SS′ as diameter touches the ellipse at real points ( i.e. at (0,b) and (0,−b)) Equation of circle will be ⇒x2+y2=a2e2=a2(a2−b2a2)=a2−b2 Circle passes through (0,b) and (0,−b) ⇒0+b2=a2−b2⇒2b2=a2⇒2a2(1−e2)=a2⇒1−e2=12⇒e2=12∴6e2=3