<p>A point on the ellipse \(x^2 + 3y^2 = 37\) where the normal is parallel to the line \(6x - 5y = 2\), is</p>
Step-by-Step Solution
Key Concept: The normal to the ellipse at point (x₀, y₀) has slope perpendicular to the tangent. For ellipse x²/a² + y²/b² = y₀, the normal slope at (x₀, y₀) is (a²y₀)/(b²x₀). Set this equal to the given slope 6/5 and use the ellipse equation to find the point.
<p><strong>Step 1:</strong> Rewrite the ellipse in standard form: x²/37 + y²/(37/3) = 1, so a² = 37, b² = 37/3</p><p><strong>Step 2:</strong> At point (x₀, y₀), the slope of tangent is found from implicit differentiation: 2x + 6y(dy/dx) = 0, giving dy/dx = -x/(3y)</p><p><strong>Step 3:</strong> The normal slope is perpendicular to tangent, so normal slope = 3y₀/x₀. The line 6x - 5y = 2 has slope 6/5</p><p><strong>Step 4:</strong> Set normal slope equal to given slope: 3y₀/x₀ = 6/5, which gives 5y₀ = 2x₀, or x₀ = 5y₀/2</p><p><strong>Step 5:</strong> Substitute into ellipse equation: (5y₀/2)² + 3y₀² = 37 → 25y₀²/4 + 3y₀² = 37 → 25y₀²/4 + 12y₀²/4 = 37 → 37y₀²/4 = 37 → y₀² = 4 → y₀ = ±2</p><p><strong>Step 6:</strong> When y₀ = 2: x₀ = 5(2)/2 = 5. When y₀ = -2: x₀ = -5</p><p><strong>Verification:</strong> 5² + 3(2)² = 25 + 12 = 37 ✓</p><p>∴ Answer: (5, 2) or (-5, -2)</p>
Correct Answer: B