Step 1: Understanding the Concept:
We are dealing with the geometry of a standard ellipse. We are given the coordinates of the foci and an endpoint of the minor axis, and a condition about the angle formed between them. We translate this geometric condition into an algebraic equation using the standard properties of an ellipse and the distance formula (or Pythagoras theorem) to solve for the eccentricity $e$.
Step 2: Key Formula or Approach:
1. Standard points on ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$: Foci $S(ae, 0)$ and $S'(-ae, 0)$. Endpoint of minor axis $B(0, b)$.
2. Right angle condition at $B$: The triangle $\Delta SBS'$ is a right-angled triangle with hypotenuse $SS'$. Thus, $BS^2 + BS'^2 = SS'^2$ (Pythagoras theorem).
3. Fundamental relationship of an ellipse: $b^2 = a^2(1 - e^2)$.
Step 3: Detailed Explanation:
Let the equation of the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.
The center is $O(0,0)$. $OB = b$ is the semi-minor axis, so the coordinates of point $B$ are $(0, b)$.
The foci are $S(ae, 0)$ and $S'(-ae, 0)$.
The angle $\angle SBS'$ is given as $90^\circ$ (a right angle).
This means the triangle formed by points $S, B,$ and $S'$ is a right-angled triangle at $B$.
According to the Pythagorean theorem:
\[ BS^2 + BS'^2 = SS'^2 \]
Let's calculate the squared distances using the distance formula:
\[ BS^2 = (ae - 0)^2 + (0 - b)^2 = a^2e^2 + b^2 \]
\[ BS'^2 = (-ae - 0)^2 + (0 - b)^2 = a^2e^2 + b^2 \]
The distance between the two foci $SS'$ is $2ae$, so:
\[ SS'^2 = (2ae)^2 = 4a^2e^2 \]
Substitute these back into the Pythagorean equation:
\[ (a^2e^2 + b^2) + (a^2e^2 + b^2) = 4a^2e^2 \]
Combine like terms:
\[ 2a^2e^2 + 2b^2 = 4a^2e^2 \]
Divide by 2:
\[ a^2e^2 + b^2 = 2a^2e^2 \]
Rearrange to isolate $b^2$:
\[ b^2 = 2a^2e^2 - a^2e^2 \]
\[ b^2 = a^2e^2 \]
Now, use the fundamental relation for ellipses, $b^2 = a^2(1 - e^2)$, and substitute it into our derived equation:
\[ a^2(1 - e^2) = a^2e^2 \]
Assuming $a \neq 0$, divide both sides by $a^2$:
\[ 1 - e^2 = e^2 \]
\[ 1 = 2e^2 \]
\[ e^2 = \frac{1}{2} \]
Since eccentricity is a positive ratio for an ellipse:
\[ e = \frac{1}{\sqrt{2}} \]
Step 4: Final Answer:
The eccentricity is $\frac{1}{\sqrt{2}}$.