Points $F$, $G$, and $H$ are collinear with $G$ between $F$ and $H$. The ellipse with foci at $G$ and $H$ lies inside the ellipse with foci at $F$ and $G$ and is tangent to it. The two ellipses have the same eccentricity $e$, and the ratio of their areas is $49$. (Recall that the eccentricity of an ellipse is $e=\frac ca$, where $c$ is the distance from the center to a focus, and $2a$ is the length of the major axis.) What is $e$?

1. Let the ellipse with foci $F,G$ have semi-major axis $a_1$ and the one with foci $G,H$ have semi-major axis $a_2$. With common eccentricity $e$, the distances from center to focus are $c_1=ea_1$ and $c_2=ea_2$.
2. The two ellipses are similar (same $b/a=\sqrt{1-e^2}$), so $(a_1/a_2)^2=49$ and $a_1=7a_2$.
3. Put the common line on the $x$-axis with the center of the large ellipse at the origin. Then $G=(c_1,0)$ and the center of the small ellipse is $(c_1+c_2,0)$. The small ellipse lies inside the large one and by symmetry they touch at the right ends of their major axes: $c_1+c_2+a_2=a_1$, i.e. $ea_1+ea_2+a_2=a_1$.
4. Dividing by $a_2$ and using $a_1=7a_2$: $7e+e+1=7$, so $e=\boxed{\dfrac34}$.