AMC 12 2025a Problem 14

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.

/* Made by Mathemagician108 */
import graph;
unitsize(0.15 inch);
draw(ellipse((0,0), 12, 4));
draw(ellipse((8,0), 4, 1.45));
pair F = (-6,0); pair G = (6,0); pair H = (10,0);
dot(F^^G^^H);
label("$F$", F, 1.5*S, p=fontsize(8pt));
label("$G$", G, 1.5*S, p=fontsize(8pt));
label("$H$", H, 1.5*S, p=fontsize(8pt));

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

\textbf{(A)}\frac35 \textbf{(B)}\frac{16}{25} \textbf{(C)}\frac45 \textbf{(D)}\frac{22}{23} \textbf{(E)}~\frac{44}{45}