2025 AMC 12A 第 14 题

Equal eccentricity makes area proportional to $a^2,$ so $\dfrac{a_1}{a_2} = \sqrt{2025} = 45$

The ellipses share focus $G$; equate their right vertices to get $a_1(1 - e) = a_2(1 + e)$

With the same eccentricity, $b = a\sqrt{1 - e^2},$ so the area $\pi a b \propto a^2.$ The area ratio $2025$ gives $\dfrac{a_1}{a_2} = \sqrt{2025} = 45,$ where $a_1, a_2$ are the semi-major axes.

Both ellipses share focus $G.$ On the large ellipse $G$ is the right focus, so its right vertex lies $a_1 - c_1$ to the right of $G.$ On the small ellipse $G$ is the left focus, so its right vertex lies $a_2 + c_2$ to the right of $G.$ Internal tangency makes these coincide: $$a_1 - c_1 = a_2 + c_2.$$

Using $c = ea,$ $a_1(1 - e) = a_2(1 + e),$ so $45(1 - e) = 1 + e,$ giving $46e = 44$ and $e = \dfrac{22}{23}.$

Thus, the correct answer is D.