Ellipse
Common Tangent to Parabola and Ellipse
Grade 11
Question:
<p><strong>Statement-1:</strong> An equation of a common tangent to the parabola \(y^2 = 16\sqrt{3}x\) and the ellipse \(2x^2 + y^2 = 4\) is \(y = 2x + 2\sqrt{3}\).</p><p><strong>Statement-2:</strong> If the line \(y = mx + \dfrac{4\sqrt{3}}{m}\), \((m \neq 0)\) is a common tangent to the parabola \(y^2 = 16\sqrt{3}x\) and the ellipse \(2x^2 + y^2 = 4\), then \(m\) satisfies \(m^4 + 2m^2 = 24\).</p>
<p>Statement-1 is true, Statement-2 is false.</p>
<p>Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.</p>
<p>Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.</p>
<p>Statement-1 is false, Statement-2 is true.</p>
Step-by-Step Solution
Key Concept: For a common tangent to exist, the line must simultaneously satisfy the tangency condition for both the parabola (using y² = 4ax form) and the ellipse (using discriminant = 0). The parabola's tangent form y = mx + a/m must match the ellipse's tangency requirement.
<p><strong>Step 1: Identify parabola tangent form</strong><br/>For parabola y² = 16√3·x, we have 4a = 16√3, so a = 4√3.<br/>Any tangent to this parabola: y = mx + 4√3/m (for m ≠ 0)</p><p><strong>Step 2: Apply ellipse tangency condition</strong><br/>Ellipse: 2x² + y² = 4 or x²/2 + y²/4 = 1<br/>For line y = mx + c to be tangent: c² = 2m² + 4<br/>Substituting c = 4√3/m:<br/>(4√3/m)² = 2m² + 4<br/>48/m² = 2m² + 4</p><p><strong>Step 3: Solve for m²</strong><br/>48 = 2m⁴ + 4m²<br/>2m⁴ + 4m² - 48 = 0<br/>m⁴ + 2m² - 24 = 0<br/>m⁴ + 2m² = 24 ✓ (Statement-2 verified)</p><p><strong>Step 4: Verify Statement-1</strong><br/>For y = 2x + 2√3, check if m = 2 satisfies m⁴ + 2m² = 24:<br/>16 + 8 = 24 ✓<br/>This line is indeed a common tangent.</p><p><strong>Conclusion:</strong> Both statements are TRUE, and Statement-2 explains Statement-1.<br/>∴ Answer: C (Both statements true; Statement-2 is correct explanation)</p>
Correct Answer: C