Problem jed Points F, G, and I are collinear with G between F and H…

The image depicts two ellipses. The larger ellipse has foci at points $F$ and $G$. The smaller ellipse has foci at points $G$ and $H$. The points $F, G, H$ are collinear. The smaller ellipse is nested inside the larger one and is internally tangent to it at a single point on the right side of the major axis.

Answer

The eccentricity is $e = \frac{44}{45}$ (Option E). Since the ellipses have the same eccentricity, they are geometrically similar, and their linear dimensions (semi-major axes) must be in a ratio equal to the square root of the ratio of their areas.

Explanation

1. Analyze the similarity of the ellipses The area of an ellipse is given by $A = \pi a b$. Since $b = a\sqrt{1-e^2}$, we can write: $A = \pi a^2 \sqrt{1-e^2}$ The area of an ellipse is proportional to the square of its semi-major axis $a$ and a factor involving eccentricity. Because both ellipses have the same eccentricity $e$, the ratio of their areas is the square of the ratio of their semi-major axes. Let $a_1$ be the semi-major axis of the large ellipse and $a_2$ be that of the small ellipse: $\frac{A_1}{A_2} = \left(\frac{a_1}{a_2}\right)^2 = 2025$ Taking the square root, we find the ratio of the linear dimensions: $\frac{a_1}{a_2} = \sqrt{2025} = 45$ The larger ellipse is exactly 45 times larger in linear scale than the smaller ellipse.

2. Define coordinates and distances Let $2c_1$ be the distance between foci $F$ and $G$, and $2c_2$ be the distance between foci $G$ and $H$. By definition of eccentricity, $c = ae$. Therefore: $c_1 = a_1 e, \quad c_2 = a_2 e$ These formulas relate the distance from the center to a focus to the semi-major axis length. From the similarity ratio, we also have: $c_1 = 45 c_2$ This means the distance $FG$ is 45 times the distance $GH$.

3. Express the tangency condition The ellipses are tangent at a point on the major axis. Let's set point $G$ as the origin $(0,0)$ on a coordinate plane. For the large ellipse (foci $F$ and $G$), the center is halfway between $F$ and $G$. The rightmost vertex is at a distance $a_1$ from the center, which is $a_1 - c_1$ from focus $G$. For the small ellipse (foci $G$ and $H$), it is tangent to the large ellipse at that same point. The rightmost vertex of the small ellipse is at a distance $a_2 + c_2$ from focus $G$. Matching these distances from point $G$: $a_1 - c_1 = a_2 + c_2$ This equation represents the geometric constraint that the right "tips" of the ellipses coincide.

4. Solve for eccentricity $e$ Substitute $c_1 = a_1 e$ and $c_2 = a_2 e$ into the geometric equation: $a_1 - a_1 e = a_2 + a_2 e$ Factor out the semi-major axes: $a_1(1-e) = a_2(1+e)$ Rearrange to find the ratio $a_1/a_2$: $\frac{a_1}{a_2} = \frac{1+e}{1-e}$ We know from Step 1 that this ratio is 45: $45 = \frac{1+e}{1-e}$ Solving for $e$: $45(1-e) = 1+e \implies 44 = 46e \implies e = \frac{44}{46} = \frac{22}{23}$ Wait, let us re-verify the tangency distance. In the large ellipse, $G$ is a focus. The distance from focus $G$ to the furthest vertex is $a_1 + c_1$. The distance from focus $G$ to the nearest vertex is $a_1 - c_1$. Based on the diagram, the ellipses are tangent at the vertex of the large ellipse furthest from $F$. Therefore, the distance from $G$ to the tangency point is $a_1 - c_1$. For the small ellipse, the tangency point is the vertex furthest from $G$, which is $a_2 + c_2$. Let's re-calculate: $a_1 - c_1 = a_2 + c_2 \implies a_1(1-e) = a_2(1+e)$ This matches the derivation above. However, if the diagram implies the tangency occurs at the vertex of the large ellipse closest to the shared focus $G$, then $a_1 - c_1 = a_2 + c_2$. If the distance from $G$ to the tangency point is determined by the small ellipse's major axis, $a_2 + c_2$ is the distance from focus $G$ to the right vertex. Given the scale $a_1 = 45 a_2$: $45a_2(1-e) = a_2(1+e) \implies 45 - 45e = 1 + e \implies 44 = 46e \implies e = \frac{22}{23}$

Value	Large Ellipse	Small Ellipse	Ratio (L/S)
Area	$A_1$	$A_2$	2025
Semi-major axis	$a_1$	$a_2$	45
Distance (Center to Focus)	$c_1$	$c_2$	45

Final Answer

Based on the geometric configuration $a_1(1-e) = a_2(1+e)$ and the similarity ratio of 45: $\boxed{\frac{22}{23}}$ (Note: If the problem assumes the center of the large ellipse is shared, the result would differ, but based on the provided focal points F, G and G, H, the tangency occurs at $x = c_1 + (a_1-c_1)$ relative to the center, leading to the derived ratio.)

Common Mistakes

• Forgetting to square root the area ratio: Students often use 2025 as the linear ratio instead of $\sqrt{2025} = 45$.
• Misidentifying the distance from focus to vertex: It is crucial to remember that the distance from a focus to the nearest vertex is $a-c = a(1-e)$ and to the furthest vertex is $a+c = a(1+e)$.