Ellipse
Eccentricity of Ellipse
Grade 11
Question:
<p>An ellipse having foci \((3, 1)\) and \((1, 1)\) passes through the point \((1, 3)\) has the eccentricity</p>
<p>(a) \(\sqrt{2} - 1\)</p>
<p>(b) \(\sqrt{3} - 1\)</p>
<p>(c) \(\dfrac{\sqrt{2} - 1}{2}\)</p>
<p>(d) \(\dfrac{\sqrt{3} - 1}{2}\)</p>
Step-by-Step Solution
Key Concept: Use the definition that sum of distances from any point on the ellipse to the two foci equals 2a, then find the distance between foci (2c) to calculate eccentricity e = c/a.
<p><strong>Step 1:</strong> Find the distance from point P(1, 3) to focus F₁(3, 1):</p><p>PF₁ = √[(1-3)² + (3-1)²] = √[4 + 4] = √8 = 2√2</p><p><strong>Step 2:</strong> Find the distance from point P(1, 3) to focus F₂(1, 1):</p><p>PF₂ = √[(1-1)² + (3-1)²] = √[0 + 4] = 2</p><p><strong>Step 3:</strong> By the definition of ellipse, sum of focal distances equals 2a:</p><p>2a = PF₁ + PF₂ = 2√2 + 2 = 2(√2 + 1)</p><p>∴ a = √2 + 1</p><p><strong>Step 4:</strong> Find the distance between foci (which equals 2c):</p><p>2c = √[(3-1)² + (1-1)²] = √4 = 2</p><p>∴ c = 1</p><p><strong>Step 5:</strong> Calculate eccentricity:</p><p>e = c/a = 1/(√2 + 1) = (√2 - 1)/[(√2 + 1)(√2 - 1)] = (√2 - 1)/(2 - 1) = √2 - 1</p><p>∴ Answer: e = √2 - 1</p>
Correct Answer: A