<p>The equation \(14x^2 - 4xy + 11y^2 - 44x - 58y + 71 = 0\) represents</p>
Step-by-Step Solution
Key Concept: Transform the general second-degree equation into standard form by completing the square after rotating axes to eliminate the xy term, then identify the conic by analyzing the discriminant B² - 4AC.
<p><strong>Step 1: Check the discriminant</strong></p><p>For equation Ax² + Bxy + Cy² + Dx + Ey + F = 0</p><p>A = 14, B = -4, C = 11</p><p>Discriminant: B² - 4AC = (-4)² - 4(14)(11) = 16 - 616 = -600 < 0</p><p>Since B² - 4AC < 0 and A ≠ C, this represents a rotated ellipse (or possibly a point or empty set).</p><p><strong>Step 2: Complete the square to verify non-degeneracy</strong></p><p>Rearrange: 14x² - 4xy + 11y² - 44x - 58y + 71 = 0</p><p>Group terms: 14(x² - xy/14) + 11(y² - 58y/11) - 44x + 71 = 0</p><p>Completing the square systematically (or using rotation to eliminate xy term):</p><p>After transformation, this reduces to an equation of the form (λ₁u² + λ₂v² = positive constant) where both λ₁, λ₂ > 0.</p><p><strong>Step 3: Conclusion</strong></p><p>The equation represents a <strong>real ellipse</strong> in the rotated coordinate system.</p><p>∴ Answer: B</p>
Correct Answer: B