Ellipse
Eccentricity of Ellipse
Grade 11
Question:
<p>The eccentricity of an ellipse having centre at the origin, axes along the coordinate axes and passing through the points \((4, -1)\) and \((-2, 2)\) is</p>
<p>\(\dfrac{\sqrt{3}}{2}\)</p>
<p>\(\dfrac{\sqrt{3}}{4}\)</p>
<p>\(\dfrac{2}{\sqrt{5}}\)</p>
<p>\(\dfrac{1}{2}\)</p>
Step-by-Step Solution
Key Concept: Use the standard ellipse equation x²/a² + y²/b² = 1 and substitute both given points to create a system of two equations in two unknowns (a² and b²), then find eccentricity e = √(1 - b²/a²).
<p><strong>Step 1:</strong> Let the ellipse be x²/a² + y²/b² = 1 (axes along coordinate axes, centre at origin).</p><p><strong>Step 2:</strong> Point (4, -1) lies on ellipse: 16/a² + 1/b² = 1 ... (i)</p><p><strong>Step 3:</strong> Point (-2, 2) lies on ellipse: 4/a² + 4/b² = 1 ... (ii)</p><p><strong>Step 4:</strong> From (i): 16/a² + 1/b² = 1. From (ii): 4/a² + 4/b² = 1. Multiply (ii) by 4: 16/a² + 16/b² = 4 ... (iii)</p><p><strong>Step 5:</strong> Subtract (i) from (iii): 16/b² - 1/b² = 4 - 1, so 15/b² = 3, thus b² = 5.</p><p><strong>Step 6:</strong> Substitute b² = 5 in (i): 16/a² + 1/5 = 1, so 16/a² = 4/5, thus a² = 20.</p><p><strong>Step 7:</strong> Since a² = 20 > b² = 5, major axis is along x-axis. e = √(1 - b²/a²) = √(1 - 5/20) = √(15/20) = √(3/4) = √3/2.</p><p>∴ Answer: A (e = √3/2)</p>
Correct Answer: A