<p>If the tangents on the ellipse \(4x^2 + y^2 = 8\) at the points \((1, 2)\) and \((a, b)\) are perpendicular to each other, then \(a^2\) is equal to __________ (up to four decimal places).</p>
Step-by-Step Solution
Key Concept: For an ellipse, the slope of the tangent at any point can be found using implicit differentiation. Two tangents are perpendicular when the product of their slopes equals -1. Use this condition along with the constraint that (a,b) lies on the ellipse to find a².
<p><strong>Step 1:</strong> Find the slope of tangent at (1, 2) using implicit differentiation on 4x² + y² = 8.</p><p>Differentiating: 8x + 2y(dy/dx) = 0 → dy/dx = -4x/y</p><p>At (1, 2): slope m₁ = -4(1)/2 = -2</p><p><strong>Step 2:</strong> Since tangents are perpendicular, m₁ · m₂ = -1.</p><p>Therefore: (-2) · m₂ = -1 → m₂ = 1/2</p><p><strong>Step 3:</strong> At point (a, b), the slope is -4a/b = 1/2.</p><p>This gives: -8a = b, so b = -8a</p><p><strong>Step 4:</strong> Since (a, b) lies on the ellipse: 4a² + b² = 8</p><p>Substitute b = -8a: 4a² + 64a² = 8</p><p>68a² = 8</p><p>a² = 8/68 = 2/17</p><p><strong>Step 5:</strong> Calculate: a² = 2/17 ≈ 0.1176</p><p>∴ Answer: <strong>0.1176</strong></p>
Correct Answer: 0