<p>An ellipse passes through the points \((4, -1)\) and \((-2, 2)\). Find the required eccentricity of the ellipse.</p>
Step-by-Step Solution
Key Concept: For an ellipse with center at origin and axes along coordinate axes, use the standard form x²/a² + y²/b² = 1. Substitute both given points to create two equations, solve for a² and b², then use e² = 1 - b²/a² (assuming a > b).
<p><strong>Step 1:</strong> Assume ellipse center at origin with equation x²/a² + y²/b² = 1</p><p><strong>Step 2:</strong> Substitute point (4, -1): 16/a² + 1/b² = 1 ... (i)</p><p><strong>Step 3:</strong> Substitute point (-2, 2): 4/a² + 4/b² = 1 ... (ii)</p><p><strong>Step 4:</strong> Let u = 1/a² and v = 1/b². Then: 16u + v = 1 and 4u + 4v = 1</p><p><strong>Step 5:</strong> From equation (i): v = 1 - 16u. Substitute into (ii): 4u + 4(1 - 16u) = 1 → 4u + 4 - 64u = 1 → -60u = -3 → u = 1/20</p><p><strong>Step 6:</strong> Therefore a² = 20 and v = 1 - 16(1/20) = 1 - 4/5 = 1/5, so b² = 5</p><p><strong>Step 7:</strong> Since a² = 20 > b² = 5, major axis is along x-axis. e² = 1 - b²/a² = 1 - 5/20 = 15/20 = 3/4</p><p><strong>Step 8:</strong> e = √(3/4) = √3/2</p><p>∴ Answer: B (e = √3/2)</p>
Correct Answer: B