Ellipse
Eccentricity
Grade 11
Question:
<p>If <i>B</i> is one end of the minor axis of an ellipse and <i>F</i><sub>1</sub>, <i>F</i><sub>2</sub> are its foci such that ∠<i>F</i><sub>1</sub><i>BF</i><sub>2</sub> = 90°, then the eccentricity of the ellipse is</p>
<p>\(\dfrac{1}{2}\)</p>
<p>\(\dfrac{1}{\sqrt{2}}\)</p>
<p>\(\dfrac{\sqrt{3}}{2}\)</p>
<p>\(\dfrac{1}{\sqrt{3}}\)</p>
Step-by-Step Solution
Key Concept: When B is at the end of the minor axis, it lies on the circle with foci as diameter. Use the angle condition ∠F1BF2 = 90° along with the relationship between semi-major axis a, semi-minor axis b, and distance between foci 2c.
<p><strong>Step 1:</strong> Let B = (0, b) be an end of the minor axis. The foci are F₁ = (c, 0) and F₂ = (-c, 0).</p><p><strong>Step 2:</strong> Calculate |BF₁|² and |BF₂|²:<br/>|BF₁|² = c² + b²<br/>|BF₂|² = c² + b²<br/>So |BF₁| = |BF₂| = √(c² + b²)</p><p><strong>Step 3:</strong> Since ∠F1BF2 = 90°, by Pythagoras theorem on triangle F₁BF₂:<br/>|BF₁|² + |BF₂|² = |F₁F₂|²<br/>(c² + b²) + (c² + b²) = (2c)²<br/>2(c² + b²) = 4c²</p><p><strong>Step 4:</strong> Simplify:<br/>c² + b² = 2c²<br/>b² = c²</p><p><strong>Step 5:</strong> Use b² = a² - c²:<br/>c² = a² - c²<br/>2c² = a²<br/>e = c/a = 1/√2 = √2/2</p><p>∴ Answer: B (e = 1/√2 or √2/2)</p>
Correct Answer: B