<p>We have \(y^2 = 16\sqrt{3}x\) and \(\dfrac{x^2}{2} + \dfrac{y^2}{4} = 1\). The line \(y = mx + \dfrac{4\sqrt{3}}{m}\) is tangent to the parabola and also tangent to the ellipse. Find \(m^2\).</p>
Step-by-Step Solution
Key Concept: A line tangent to both curves must satisfy the tangency condition for the parabola (discriminant = 0) AND the ellipse (distance from center equals specific form). Use the standard tangent forms: for parabola y² = 4ax, tangent is y = mx + a/m, and for ellipse, tangency requires a²m² + b² = c².
<p><strong>Step 1:</strong> Identify the parabola form. For y² = 16√3·x, we have 4a = 16√3, so a = 4√3.</p><p><strong>Step 2:</strong> Verify the given line is tangent to parabola. The standard tangent to y² = 4ax is y = mx + a/m. Here, y = mx + 4√3/m matches perfectly, confirming tangency to the parabola for any m ≠ 0.</p><p><strong>Step 3:</strong> Apply ellipse tangency condition. For ellipse x²/a² + y²/b² = 1 with a² = 2, b² = 4, the line y = mx + c is tangent when: a²m² + b² = c²</p><p><strong>Step 4:</strong> Substitute into ellipse tangency. Here c = 4√3/m, so:</p><p>2m² + 4 = (4√3/m)²</p><p>2m² + 4 = 48/m²</p><p><strong>Step 5:</strong> Multiply by m²: 2m⁴ + 4m² = 48</p><p>2m⁴ + 4m² - 48 = 0</p><p>m⁴ + 2m² - 24 = 0</p><p><strong>Step 6:</strong> Let u = m². Then u² + 2u - 24 = 0</p><p>(u + 6)(u - 4) = 0</p><p>So u = 4 (since u = m² ≥ 0)</p><p>∴ <strong>m² = 4</strong></p>
Correct Answer: A