Ellipse
Eccentricity
Grade None
Question:
<p>If the distance between the foci of an ellipse is half the length of its latus rectum, then the eccentricity of the ellipse is</p>
<p>\(\dfrac{1}{2}\)</p>
<p>\(\dfrac{2\sqrt{2}-1}{2}\)</p>
<p>\(\sqrt{2}-1\)</p>
<p>\(\dfrac{\sqrt{2}-1}{2}\)</p>
Step-by-Step Solution
Key Concept: The latus rectum of an ellipse is 2b²/a, and the distance between foci is 2ae. Setting up the equation 2ae = ½(2b²/a) and using b² = a²(1-e²) directly yields the eccentricity.
<p><strong>Step 1:</strong> Write the given condition. Distance between foci = ½ × Latus rectum</p><p>2ae = ½ × (2b²/a)</p><p><strong>Step 2:</strong> Simplify the equation</p><p>2ae = b²/a</p><p>2a²e = b²</p><p><strong>Step 3:</strong> Use b² = a²(1 - e²)</p><p>2a²e = a²(1 - e²)</p><p>2e = 1 - e²</p><p>e² + 2e - 1 = 0</p><p><strong>Step 4:</strong> Solve using quadratic formula</p><p>e = (-2 ± √(4 + 4))/2 = (-2 ± 2√2)/2 = -1 ± √2</p><p>Since 0 < e < 1 for an ellipse: e = √2 - 1</p><p>∴ Answer: C (√2 - 1)</p>
Correct Answer: C