<p>\(P\) and \(Q\) are two distinct points on the parabola, \(y^2 = 4x\), with parameters \(t\) and \(t_1\), respectively. If the normal at \(P\) passes through \(Q\), then the minimum value of \(t_1^2\) is</p>
Step-by-Step Solution
Key Concept: The normal at point P(t²,2t) on parabola y²=4x has equation y=-tx+2t+t³. For this normal to pass through Q(t₁²,2t₁), we get a relation between t and t₁, then minimize t₁² using calculus or AM-GM.
<p><strong>Step 1: Write coordinates using parameter</strong></p><p>Point P on y²=4x with parameter t: P(t², 2t)</p><p>Point Q on y²=4x with parameter t₁: Q(t₁², 2t₁)</p><p><strong>Step 2: Find equation of normal at P</strong></p><p>For parabola y²=4x, slope of tangent at P is dy/dx = 1/t</p><p>Slope of normal = -t</p><p>Normal equation: y - 2t = -t(x - t²)</p><p>Simplifying: y = -tx + t³ + 2t</p><p><strong>Step 3: Apply condition that normal passes through Q</strong></p><p>Substitute Q(t₁², 2t₁) into normal equation:</p><p>2t₁ = -t·t₁² + t³ + 2t</p><p>2t₁ - 2t = -t·t₁² + t³</p><p>2(t₁ - t) = t(t² - t₁²)</p><p>2(t₁ - t) = t(t - t₁)(t + t₁)</p><p>Since P and Q are distinct, t₁ ≠ t, divide by (t₁ - t):</p><p>2 = -t(t + t₁)</p><p>t₁ = -2/t - t</p><p><strong>Step 4: Minimize t₁²</strong></p><p>t₁² = (-2/t - t)² = (2/t + t)²</p><p>Let f(t) = (2/t + t)² where t > 0 (considering t ≠ 0)</p><p>By AM-GM: 2/t + t ≥ 2√(2/t · t) = 2√2</p><p>Equality when 2/t = t, so t² = 2, giving t = √2</p><p>Minimum value of (2/t + t)² = (2√2)² = 8</p><p>∴ Answer: A</p>
Correct Answer: A