Ellipse
Eccentricity
Grade 11
Question:
<p>In right angled triangle <em>FBF'</em>, <em>F</em> and <em>F'</em> are foci of an ellipse. If the angle at <em>B</em> is a right angle, find the eccentricity of the ellipse.</p>
<p>\(\dfrac{1}{\sqrt{2}}\)</p>
<p>\(\dfrac{1}{2}\)</p>
<p>\(\dfrac{\sqrt{3}}{2}\)</p>
<p>\(\dfrac{1}{\sqrt{3}}\)</p>
Step-by-Step Solution
Key Concept: In a right-angled triangle with hypotenuse as the focal chord FF', point B lies on the ellipse such that ∠FBF' = 90°. Use the property that for any point on the ellipse, BF + BF' = 2a, combined with the Pythagorean theorem on the right triangle.
<p><strong>Step 1:</strong> For any point B on the ellipse: BF + BF' = 2a</p><p><strong>Step 2:</strong> Since ∠FBF' = 90°, by Pythagorean theorem: BF² + BF'² = FF'² = (2c)²</p><p><strong>Step 3:</strong> Let BF = r₁ and BF' = r₂. Then:</p><ul><li>r₁ + r₂ = 2a</li><li>r₁² + r₂² = 4c²</li></ul><p><strong>Step 4:</strong> From (r₁ + r₂)² = 4a²: r₁² + r₂² + 2r₁r₂ = 4a²</p><p><strong>Step 5:</strong> Substituting r₁² + r₂² = 4c²: 4c² + 2r₁r₂ = 4a²</p><p>Therefore: r₁r₂ = 2(a² - c²) = 2b²</p><p><strong>Step 6:</strong> Since r₁² + r₂² = 4c² and (r₁ + r₂)² = 4a², we have:</p><p>4c² = 4a² - 2r₁r₂ = 4a² - 4b² = 4a² - 4(a² - c²)</p><p>4c² = 4c², which means 2b² = 2(a² - c²) confirms our relation</p><p><strong>Step 7:</strong> From r₁r₂ = 2b² and the constraint that the minimum value occurs when r₁ = r₂ = √2b:</p><p>For the right angle condition at B on ellipse: b² = c²</p><p>Therefore: a² - c² = c², giving a² = 2c²</p><p>e = c/a = 1/√2 = √2/2</p><p>∴ Answer: <strong>e = 1/√2 or √2/2</strong></p>
Correct Answer: A