Question

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 $e?$

$\dfrac{3}{5}$

$\dfrac{16}{25}$

$\dfrac{4}{5}$

$\dfrac{22}{23}$

$\dfrac{44}{45}$

Solution:

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 Problem 14

Solution:

Problem 14 in Other Years