<p>Given the equation of parabola \((y-2)^2 = (x-1)\), find the area bounded by the parabola and its tangent at \(P(2, 3)\) and the line \(x = 2y - 4\).<br>\[\Delta = \int_0^3 \left[(y-2)^2 + 1 - (2y - 4)\right] dy\]</p>
Step-by-Step Solution
Key Concept: The area is bounded by three curves: the parabola (y-2)² = x-1, a tangent line at P(2,3), and a chord x = 2y-4. Setting up the integral requires expressing x in terms of y and identifying correct limits by finding intersection points.
<p><strong>Step 1: Verify point P(2,3) lies on parabola</strong></p><p>Substitute into (y-2)² = x-1: (3-2)² = 2-1 ⟹ 1 = 1 ✓</p><p><strong>Step 2: Find tangent line at P(2,3)</strong></p><p>From (y-2)² = x-1, differentiating: 2(y-2)dy = dx ⟹ dx/dy = 2(y-2)</p><p>At P(2,3): dx/dy = 2(3-2) = 2</p><p>Tangent line: x - 2 = 2(y - 3) ⟹ x = 2y - 4</p><p><strong>Step 3: Identify the bounded region</strong></p><p>The parabola x = (y-2)² + 1 and the line x = 2y - 4 intersect where:</p><p>(y-2)² + 1 = 2y - 4</p><p>y² - 4y + 4 + 1 = 2y - 4</p><p>y² - 6y + 9 = 0 ⟹ (y-3)² = 0 ⟹ y = 3 (tangent point)</p><p>For the other boundary, check where the line intersects the x-axis region. When y = 0: x = -4 and parabola gives x = 5. The region is bounded between y = 0 and y = 3.</p><p><strong>Step 4: Set up and evaluate the integral</strong></p><p>Area = ∫₀³ [(2y - 4) - ((y-2)² + 1)] dy</p><p>= ∫₀³ [2y - 4 - y² + 4y - 4 - 1] dy</p><p>= ∫₀³ [-y² + 6y - 9] dy</p><p>= ∫₀³ [-(y² - 6y + 9)] dy</p><p>= ∫₀³ [-(y-3)²] dy</p><p>= [-⅓(y-3)³]₀³</p><p>= [-⅓(0)] - [-⅓(-27)]</p><p>= 0 - 9 = -9</p><p>Taking absolute value: |−9| = <strong>9</strong></p>
Correct Answer: 9