The area bounded by the curves y² = 4a(x + a) and y² = 4b(b − x), where a, b > 0, is:

The area bounded by the curves y² = 4a(x + a) and y² = 4b(b − x), where a, b > 0, is:

• A. (8/3)√(ab)(a + b) sq units
• B. (4/3)(a + b)√(ab) sq units
• C. (2/3)√(ab)(a + b) sq units
• D. 4√(ab)(a + b) sq units

Answer: A) (8/3)√(ab)(a + b) sq units

Explanation: First: y² = 4a(x+a) → x = y²/(4a) − a. Second: y² = 4b(b−x) → x = b − y²/(4b). Intersection: y²/(4a) − a = b − y²/(4b) → y²/(4a) + y²/(4b) = a + b → y²(1/(4a) + 1/(4b)) = a+b → y²((a+b)/(4ab)) = a+b → y² = 4ab → y = ±2√(ab). Area = ∫[−2√(ab) to 2√(ab)] [(b − y²/(4b)) − (y²/(4a) − a)] dy = ∫ (a+b − y²/(4)(1/a+1/b)) dy = 2∫[0 to 2√(ab)] [a+b − y²(a+b)/(4ab)] dy = 2(a+b)[y − y³/(12ab)]₀^(2√(ab)) = 2(a+b)[2√(ab) − (8ab√(ab))/(12ab)] = 2(a+b)[2√(ab) − (2/3)√(ab)] = 2(a+b)(4/3)√(ab) = (8/3)(a+b)√(ab) sq units.