If OB is the semi-minor axis of an ellipse, F1 and F2 are its foci and the angle between F1B and F2B is a right angle, then the square of the eccentricity of the ellipse is

Let the equation of the ellipse be x²/a² + y²/b² = 1.
The foci are at (±ae, 0), where e is the eccentricity.
The semi-minor axis is OB = b.
The coordinates of B are (0, b).
F1 = (-ae, 0) and F2 = (ae, 0).
The slope of F1B is m1 = (b - 0)/(0 - (-ae)) = b/ae.
The slope of F2B is m2 = (b - 0)/(0 - ae) = -b/ae.
Since the angle between F1B and F2B is a right angle, the product of their slopes is -1.
m1 * m2 = (b/ae) * (-b/ae) = -b²/a²e² = -1
b² = a²e²
We know that b² = a²(1 - e²).
Therefore, a²(1 - e²) = a²e²
1 - e² = e²
2e² = 1
e² = 1/2
Thus, the square of the eccentricity is 1/2.