<p>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</p>
Step-by-Step Solution
Key Concept: When B is at the end of the semi-minor axis and ∠F₁BF₂ = 90°, use the property that BF₁² + BF₂² = F₁F₂² (Pythagorean theorem) combined with the focal chord property BF₁ + BF₂ = 2a to find eccentricity.
<p><strong>Step 1:</strong> Set up coordinates with center O at origin. Let B = (0, b) be the end of semi-minor axis. Foci are F₁ = (-c, 0) and F₂ = (c, 0) where c² = a² - b².</p><p><strong>Step 2:</strong> Since ∠F₁BF₂ = 90°, we have BF₁² + BF₂² = F₁F₂² = (2c)² = 4c²</p><p><strong>Step 3:</strong> Calculate: BF₁² = c² + b² and BF₂² = c² + b², so BF₁² + BF₂² = 2(c² + b²) = 4c²</p><p><strong>Step 4:</strong> This gives 2(c² + b²) = 4c², so 2b² = 2c², thus b² = c²</p><p><strong>Step 5:</strong> Since b² = a² - c², we have c² = a² - c², which gives 2c² = a²</p><p><strong>Step 6:</strong> Therefore e² = c²/a² = a²/2a² = 1/2</p><p>∴ Answer: A (e² = 1/2)</p>
Correct Answer: A