Parabola
Common Tangent
Grade 11
Question:
<p>If the line \(ax + y + c\), touches both the curves \(x^2 + y^2 = 1\) and \(y^2 = 4\sqrt{2}x\), then \(|c|\) is equal to __________ (up to three decimal places).</p>
Step-by-Step Solution
Key Concept: A line is tangent to both curves simultaneously when it satisfies the tangency condition for the circle (distance from center equals radius) AND the tangency condition for the parabola (discriminant = 0). These two independent conditions determine the relationship between a and c.
<p><strong>Step 1: Tangency condition for circle x² + y² = 1</strong></p><p>Line: ax + y + c = 0. Distance from origin to line = radius.</p><p>Distance = |c|/√(a² + 1) = 1</p><p>Therefore: c² = a² + 1 ... (i)</p><p><strong>Step 2: Tangency condition for parabola y² = 4√2·x</strong></p><p>From line: y = -ax - c. Substitute into y² = 4√2·x:</p><p>(-ax - c)² = 4√2·x</p><p>a²x² + 2acx + c² = 4√2·x</p><p>a²x² + (2ac - 4√2)x + c² = 0</p><p>For tangency, discriminant = 0:</p><p>(2ac - 4√2)² - 4a²c² = 0</p><p>4a²c² - 16√2·ac + 32 - 4a²c² = 0</p><p>-16√2·ac + 32 = 0</p><p>ac = 2√2/√2 = √2 ... (ii)</p><p><strong>Step 3: Solve system of equations (i) and (ii)</strong></p><p>From (ii): a = √2/c</p><p>Substitute into (i): c² = 2/c² + 1</p><p>c⁴ = 2 + c²</p><p>c⁴ - c² - 2 = 0</p><p>(c² - 2)(c² + 1) = 0</p><p>c² = 2 (since c² + 1 > 0)</p><p>|c| = √2 ≈ 1.414</p><p><strong>∴ Answer: 1.414</strong></p>
Correct Answer: 1