Parabola
Tangent to Parabola
Grade 11
Question:
<p>Let \(P\) be a point on the parabola, \(x^2 = 4y\). If the distance of \(P\) from the centre of the circle, \(x^2 + y^2 + 6x + 8 = 0\) is minimum, then the equation of the tangent to the parabola at \(P\), is</p>
<p>\(x + 4y - 2 = 0\)</p>
<p>\(x - y + 3 = 0\)</p>
<p>\(x + y + 1 = 0\)</p>
<p>\(x + 2y = 0\)</p>
Step-by-Step Solution
Key Concept: The point P on the parabola closest to the circle's center lies where the normal to the parabola passes through that center. Use parametric form and minimize distance to find the point, then write the tangent equation.
<p><strong>Step 1:</strong> Rewrite the circle equation: x² + y² + 6x + 8 = 0 → (x+3)² + y² = 1. Center C = (-3, 0), radius = 1.</p><p><strong>Step 2:</strong> Use parametric form for parabola x² = 4y: Point P = (2t, t²). Distance from P to C is: d = √[(2t+3)² + t⁴]</p><p><strong>Step 3:</strong> Minimize d² = (2t+3)² + t⁴. Taking derivative: d(d²)/dt = 2(2t+3)·2 + 4t³ = 8t + 12 + 4t³ = 0 → t³ + 2t + 3 = 0 → (t+1)(t²-t+3) = 0</p><p><strong>Step 4:</strong> Since t² - t + 3 has no real roots (discriminant = 1-12 < 0), we get t = -1. Thus P = (-2, 1).</p><p><strong>Step 5:</strong> For parabola x² = 4y, the tangent at point (x₀, y₀) is: x·x₀ = 2(y + y₀) → x(-2) = 2(y + 1) → -2x = 2y + 2 → x + y + 1 = 0</p><p>∴ Answer: C (The equation is x + y + 1 = 0)</p>
Correct Answer: C