Parabola
Common Tangents
Grade 11
Question:
<p>If the common tangents to the parabola, \(x^2 = 4y\) and the circle, \(x^2 + y^2 = 4\) intersect at the point \(P\), then the distance of \(P\) from the origin, is</p>
<p>\(2(\sqrt{2}+1)\)</p>
<p>\(3+2\sqrt{2}\)</p>
<p>\(2(3+2\sqrt{2})\)</p>
<p>\(\sqrt{2}+1\)</p>
Step-by-Step Solution
Key Concept: A common tangent to both curves must satisfy the tangency condition for the parabola (discriminant = 0) and the circle (distance from center = radius). Finding the intersection point of these common tangents requires solving the system of tangent equations simultaneously.
<p><strong>Step 1: Find tangents to parabola x² = 4y</strong></p><p>A line y = mx + c is tangent to x² = 4y if x² = 4(mx + c) has equal roots.</p><p>This gives: x² - 4mx - 4c = 0 has discriminant = 0</p><p>16m² + 16c = 0 ⟹ c = -m²</p><p>So tangents to parabola: y = mx - m²</p><p><strong>Step 2: Find tangents to circle x² + y² = 4</strong></p><p>For line y = mx + c to be tangent to circle: |c|/√(1+m²) = 2</p><p>|c| = 2√(1+m²)</p><p><strong>Step 3: Find common tangents</strong></p><p>Both conditions must hold: c = -m² and |c| = 2√(1+m²)</p><p>|-m²| = 2√(1+m²)</p><p>m⁴ = 4(1+m²)</p><p>m⁴ - 4m² - 4 = 0</p><p>m² = (4 ± √(16+16))/2 = (4 ± 4√2)/2 = 2 ± 2√2</p><p>Taking m² = 2 + 2√2 (the valid solution)</p><p><strong>Step 4: Find intersection point P of common tangents</strong></p><p>The two common tangents are: y = m₁x - m₁² and y = m₂x - m₂²</p><p>By symmetry, P lies on y-axis at x = 0</p><p>From tangent to parabola with slope m: tangent line is y = mx - m²</p><p>For the pair of common tangents with slopes ±m where m² = 2 + 2√2:</p><p>They intersect where: mx - m² = -mx - m² gives x = 0</p><p>At x = 0: y = -m² = -(2 + 2√2) = -2(1 + √2)</p><p>P = (0, -2(1+√2))</p><p>Distance from origin = |y| = 2(1 + √2) = 2 + 2√2</p><p>∴ Answer: <strong>2 + 2√2</strong> or <strong>2(1 + √2)</strong></p>
Correct Answer: A