What is the eccentricity of the ellipse if the angle between the straight lines joining the foci to an extremity of the minor axis is 90° ?

Explanation

Let an extremity of the minor axis be B(0,b) and the foci be S(ae,0) and S'(-ae,0). Since \angle SBS' = 90^{\circ}, the triangle is a right-angled isosceles triangle. The slope of BS is \frac{-b}{ae} and the slope of BS' is \frac{-b}{-ae}. Since they are perpendicular, (\frac{-b}{ae})(\frac{b}{ae}) = -1 \implies b^2 = a^2e^2. Using b^2 = a^2(1-e^2), we have a^2e^2 = a^2(1-e^2) \implies 2e^2 = 1 \implies e = \frac{1}{\sqrt{2}}.

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