Solution

The foci are at: $$S(ae,0), \quad S'(-ae,0)$$ and the end of the minor axis is: $$B(0,b), \quad \text{where } b^2 = a^2(1-e^2).$$

In an equilateral triangle ∆BSS′: $$BS = SS'.$$

Now, $$BS = \sqrt{(ae)^2 + b^2}, \quad SS' = 2ae.$$ Hence, $$\sqrt{a^2e^2 + b^2} = 2ae.$$

But, $$b^2 = a^2(1-e^2).$$

So, $$\sqrt{a^2e^2 + a^2(1-e^2)} = 2ae,$$ $$\sqrt{a^2} = 2ae,$$ $$a = 2ae \;\;\Rightarrow\;\; e = \tfrac{1}{2}.$$