<p>An ellipse has \(OB\) as semi minor axis, \(F\) and \(F'\) its foci and the angle \(FBF'\) is a right angle. Then the eccentricity of the ellipse is</p>
Step-by-Step Solution
Key Concept: When B is an endpoint of the semi-minor axis and ∠FBF' = 90°, use the property that F and F' are symmetric about the center O, combined with the right angle condition to relate a, b, and c.
<p><strong>Step 1:</strong> Set up the geometry. Let O be the center, B be an endpoint of semi-minor axis (so OB = b), and F, F' be foci with OF = OF' = c.</p><p><strong>Step 2:</strong> In triangle FBF', since F and F' are symmetric about O on the major axis, and B is on the minor axis, we have FB = F'B (by symmetry).</p><p><strong>Step 3:</strong> Since ∠FBF' = 90°, triangle FBF' is a right isosceles triangle. Using the distance formula: FF' = 2c (distance between foci).</p><p><strong>Step 4:</strong> For right triangle FBF' with ∠FBF' = 90° and FB = F'B: By Pythagorean theorem, (2c)² = FB² + F'B² = 2(FB)²</p><p><strong>Step 5:</strong> Calculate FB using coordinates. B = (0, b), F = (c, 0): FB² = c² + b². So 4c² = 2(c² + b²), giving 4c² = 2c² + 2b²</p><p><strong>Step 6:</strong> Simplify: 2c² = 2b² → c² = b². Since b² = a² - c², we get c² = a² - c² → 2c² = a² → e = c/a = 1/√2 = √2/2</p><p>∴ Answer: A (eccentricity = 1/√2 or √2/2)</p>
Correct Answer: A