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(A) Let the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with $b=2$. The vertices are $(\pm a, 0)$. Let one vertex be $(a, 0)$.Let the other two

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$\frac{\sqrt{3}}{2}$

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(A) Let the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with $b=2$. The vertices are $(\pm a, 0)$. Let one vertex be $(a, 0)$.Let the other two vertices be $(-a \cos 	heta, 2 \sin 	heta)$ and $(-a \cos 	heta, -2 \sin 	heta)$.The base of the triangle is $4 \sin 	heta$ and the height is $a + a \cos 	heta = a(1 + \cos 	heta)$.The area $A = \frac{1}{2} 	imes (4 \sin 	heta) 	imes a(1 + \cos 	heta) = 2a \sin 	heta (1 + \cos 	heta)$.To maximize $A$,let $f(	heta) = \sin 	heta (1 + \cos 	heta)$.$f'(	heta) = \cos 	heta (1 + \cos 	heta) - \sin^2 	heta = \cos 	heta + \cos^2 	heta - (1 - \cos^2 	heta) = 2 \cos^2 	heta + \cos 	heta - 1 = 0$.$(2 \cos 	heta - 1)(\cos 	heta + 1) = 0$.Since $	heta 
eq \pi$,we have $\cos 	heta = \frac{1}{2}$,so $\sin 	heta = \frac{\sqrt{3}}{2}$.$A_{\max} = 2a \left(\frac{\sqrt{3}}{2}ight) \left(1 + \frac{1}{2}ight) = 2a \left(\frac{\sqrt{3}}{2}ight) \left(\frac{3}{2}ight) = \frac{3\sqrt{3}}{2} a$.Given $A_{\max} = 6\sqrt{3}$,so $\frac{3\sqrt{3}}{2} a = 6\sqrt{3} \Rightarrow a = 4$.Eccentricity $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{4}{16}} = \sqrt{1 - \frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$.

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