If OB is the semi-minor axis of an ellipse, `F_1` and `F_2` are its foci and the angle between `F_1B` and `F_2B` is a right angle, then the square of the eccentricity of the ellipse is

Options

A. `1/2`
B. `1/sqrt2`
C. `1/(2sqrt2)`
D. `1/4`

Answer

To solve the problem, we need to find the square of the eccentricity of the ellipse given that the angle between the lines from the foci \( F_1 \) and \( F_2 \) to the point \( B \) (which is on the semi-minor axis) is a right angle.

### Step-by-Step Solution:

1. **Understanding the Ellipse**:
The standard form of the equation of an ellipse centered at the origin is given by:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]
where \( a \) is the semi-major axis and \( b \) is the semi-minor axis. Here, \( OB = b \).

2. **Coordinates of the Foci**:
The foci \( F_1 \) and \( F_2 \) of the ellipse are located at:
\[ F_1 = (-ae, 0) \quad \text{and} \quad F_2 = (ae, 0) \]
where \( e \) is the eccentricity of the ellipse.

3. **Coordinates of Point B**:
Since \( B \) lies on the semi-minor axis, its coordinates are:
\[ B = (0, b) \]

4. **Using the Right Angle Condition**:
The angle between the lines \( F_1B \) and \( F_2B \) being a right angle implies that the vectors \( \overrightarrow{F_1B} \) and \( \overrightarrow{F_2B} \) are perpendicular.

The vectors are:
\[ \overrightarrow{F_1B} = (0 - (-ae), b - 0) = (ae, b) \]
\[ \overrightarrow{F_2B} = (0 - (ae), b - 0) = (-ae, b) \]

For these two vectors to be perpendicular, their dot product must equal zero:
\[ (ae)(-ae) + (b)(b) = 0 \]
Simplifying this gives:
\[ -a^2 e^2 + b^2 = 0 \]
Thus:
\[ b^2 = a^2 e^2 \]

5. **Relating b² to a² and e²**:
We know from the properties of ellipses that:
\[ b^2 = a^2(1 - e^2) \]
Setting the two expressions for \( b^2 \) equal gives:
\[ a^2 e^2 = a^2(1 - e^2) \]
Dividing both sides by \( a^2 \) (assuming \( a \neq 0 \)):
\[ e^2 = 1 - e^2 \]

6. **Solving for e²**:
Rearranging gives:
\[ 2e^2 = 1 \quad \Rightarrow \quad e^2 = \frac{1}{2} \]

### Final Answer:
The square of the eccentricity of the ellipse is:
\[ \boxed{\frac{1}{2}} \]