<p>Let \(y^2 = x\) be a given parabola and a variable chord cuts the parabola at \(P\) and \(Q\). Let \(C\) be the vertex of parabola. If the locus of the point of intersection of tangents at \(P\) and \(Q\) is \(x + 1 = 0\), then the minimum area of triangle \(PCQ\) be \(M\). Find \(4M\).</p>
Step-by-Step Solution
Key Concept: For parabola y² = x, if tangents at points P and Q meet at a point on directrix x = -1/4, then the chord PQ passes through focus. The area of triangle PCQ is minimized when the chord is perpendicular to the axis, making it the latus rectum.
<p><strong>Step 1:</strong> For parabola y² = x, we have 4a = 1, so a = 1/4. Focus is at F(1/4, 0) and directrix is x = -1/4.</p><p><strong>Step 2:</strong> Let P(t₁², t₁) and Q(t₂², t₂) be two points on the parabola. The equation of tangent at P is: t₁y = x + t₁². Similarly at Q: t₂y = x + t₂².</p><p><strong>Step 3:</strong> These tangents intersect where t₁y - t₁² = t₂y - t₂², giving y(t₁ - t₂) = t₁² - t₂². Thus y = t₁ + t₂ and substituting: x = t₁t₂.</p><p><strong>Step 4:</strong> Given the locus of intersection is x = -1, we have t₁t₂ = -1 (interpreting the given condition as applying to the chord configuration).</p><p><strong>Step 5:</strong> Vertex C is at origin (0, 0). Points are P(t₁², t₁) and Q(t₂², t₂) with t₁t₂ = -1.</p><p><strong>Step 6:</strong> Area of triangle PCQ = (1/2)|t₁² · t₂ - t₂² · t₁| = (1/2)|t₁t₂||t₁ - t₂| = (1/2)|t₁ - t₂|</p><p><strong>Step 7:</strong> Since t₁t₂ = -1, let t₁ = t and t₂ = -1/t. Then |t₁ - t₂| = |t + 1/t| ≥ 2 (by AM-GM), with minimum when t = 1 (or t = -1).</p><p><strong>Step 8:</strong> Minimum area M = (1/2) · 2 = 1</p><p><strong>∴ Answer: 4M = 4(1) = 4</strong></p>
Correct Answer: 4