<p>The eccentric angle of a point on the ellipse \(x^2 + 3y^2 = 6\) at a distance 2 units from the centre of the ellipse is</p>
Step-by-Step Solution
Key Concept: Convert the ellipse to standard form to identify semi-major and semi-minor axes, then use the parametric form (a·cosθ, b·sinθ) where the distance from center equals 2 to find the eccentric angle θ.
<p><strong>Step 1:</strong> Convert to standard form: x²/6 + y²/2 = 1, so a² = 6, b² = 2, giving a = √6, b = √2</p><p><strong>Step 2:</strong> Parametric form of ellipse: x = √6·cosθ, y = √2·sinθ where θ is the eccentric angle</p><p><strong>Step 3:</strong> Distance from center: x² + y² = 2²<br/>6cos²θ + 2sin²θ = 4<br/>6cos²θ + 2(1-cos²θ) = 4<br/>6cos²θ + 2 - 2cos²θ = 4<br/>4cos²θ = 2<br/>cos²θ = 1/2</p><p><strong>Step 4:</strong> Therefore cosθ = ±1/√2, which gives θ = π/4, 3π/4, 5π/4, or 7π/4</p><p>∴ Answer: A (θ = π/4 or 3π/4, depending on which quadrant)</p>
Correct Answer: A