<p>If the tangent to the parabola \(y^2 = x\) at a point \((\alpha, \beta)\), \((\beta > 0)\) is also a tangent to the ellipse, \(x^2 + 2y^2 = 1\), then \(\alpha\) is equal to __________ (up to three decimal places).</p>
Step-by-Step Solution
Key Concept: The tangent line to the parabola y² = x at point (α, β) must satisfy the parabola's tangent equation, and then apply the condition that this same line is tangent to the ellipse (discriminant = 0).
<p><strong>Step 1:</strong> Since (α, β) lies on parabola y² = x, we have β² = α.</p><p><strong>Step 2:</strong> The tangent to y² = x at (α, β) is: βy = ½(x + α). Substituting α = β²: <strong>βy = ½(x + β²)</strong> or <strong>2βy - x - β² = 0</strong>.</p><p><strong>Step 3:</strong> For this line to be tangent to ellipse x² + 2y² = 1, the distance from center (0,0) to the line must equal the semi-axis length in that direction. Using the tangency condition:</p><p>From line: x = 2βy - β²</p><p>Substitute into ellipse: (2βy - β²)² + 2y² = 1</p><p>4β²y² - 4β³y + β⁴ + 2y² = 1</p><p>(4β² + 2)y² - 4β³y + (β⁴ - 1) = 0</p><p><strong>Step 4:</strong> For tangency, discriminant Δ = 0:</p><p>(4β³)² - 4(4β² + 2)(β⁴ - 1) = 0</p><p>16β⁶ - 4(4β⁶ - 4β² + 2β⁴ - 2) = 0</p><p>16β⁶ - 16β⁶ + 16β² - 8β⁴ + 8 = 0</p><p>-8β⁴ + 16β² + 8 = 0</p><p>β⁴ - 2β² - 1 = 0</p><p><strong>Step 5:</strong> Let u = β²: u² - 2u - 1 = 0</p><p>u = (2 ± √(4 + 4))/2 = (2 ± 2√2)/2 = 1 ± √2</p><p>Since β > 0, we need β² = 1 + √2 (taking positive root)</p><p>Therefore: <strong>α = β² = 1 + √2 ≈ 2.414</strong></p><p>∴ Answer: <strong>2.414</strong></p>
Correct Answer: 2