Question

点 $F$ 、 $G$ 和 $H$ 共线，且 $G$ 在 $F$ 和 $H$ 之间。以 $G$ 和 $H$ 为焦点的椭圆与以 $F$ 和 $G$ 为焦点的椭圆内切，如下图所示。

这两个椭圆的离心率相同，均为 $e$ ，且它们的面积之比为 $2025$ 。（回忆：椭圆的离心率为 $e = \dfrac{c}{a}$ ，其中 $c$ 是从中心到焦点的距离， $2a$ 是长轴长度。） $e$ 是多少？

Points $F,$ $G,$ and $H$ are collinear with $G$ between $F$ and $H.$ The ellipse with foci at $G$ and $H$ is internally tangent to the ellipse with foci at $F$ and $G,$ as shown below.

The two ellipses have the same eccentricity $e,$ and the ratio of their areas is $2025.$ (Recall that the eccentricity of an ellipse is $e = \dfrac{c}{a},$ where $c$ is the distance from the center to a focus, and $2a$ is the length of the major axis.) What is $e?$

$\dfrac{3}{5}$

$\dfrac{16}{25}$

$\dfrac{4}{5}$

$\dfrac{22}{23}$

$\dfrac{44}{45}$

答案：D

知识点：椭圆面积比

难度评级：1730

解答：

离心率相同，则 $b = a\sqrt{1 - e^2}$ ，所以面积 $\pi a b \propto a^2$ 。面积之比为 $2025$ ，因此 $\dfrac{a_1}{a_2} = \sqrt{2025} = 45$ ，其中 $a_1, a_2$ 是两个半长轴。

两个椭圆共用焦点 $G$ 。在大椭圆中， $G$ 是右焦点，所以右顶点位于 $G$ 右侧 $a_1 - c_1$ 处。在小椭圆中， $G$ 是左焦点，所以右顶点位于 $G$ 右侧 $a_2 + c_2$ 处。内切使这两个点重合： $a_1 - c_1 = a_2 + c_2.$

利用 $c = ea$ ，得 $a_1(1 - e) = a_2(1 + e)$ ，所以 $45(1 - e) = 1 + e$ ，从而 $46e = 44$ ， $e = \dfrac{22}{23}$ 。

因此，正确答案是 D。

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.

2025 AMC 12A 第 14 题

答案：D

难度评级：1730

解答：

其他年份的第 14 题