Parabola
Maximum Area of Triangle
Grade 11
Question:
<p>Let A(4, −4) and B(9, 6) be points on the parabola, \(y^2 = 4x\). Let C be chosen on the arc AOB of the parabola, where O is the origin, such that the area of \(\triangle ACB\) is maximum. Then, the area (in sq. units) of \(\triangle ACB\), is __________ (up to two decimal places).</p>
Step-by-Step Solution
Key Concept: For a parabola, the area of triangle ACB with fixed points A and B is maximized when C is positioned such that the perpendicular distance from C to line AB is maximum. This occurs at the point where the tangent to the parabola is parallel to AB.
<p><strong>Step 1:</strong> Verify A(4, −4) and B(9, 6) are on y² = 4x: (−4)² = 16 = 4(4) ✓ and 6² = 36 = 4(9) ✓</p><p><strong>Step 2:</strong> Find equation of line AB. Slope = (6−(−4))/(9−4) = 10/5 = 2. Line AB: y + 4 = 2(x − 4) → y = 2x − 12</p><p><strong>Step 3:</strong> For maximum area, find C on parabola where tangent is parallel to AB (slope = 2). For y² = 4x, dy/dx = 2/y. Set 2/y = 2 → y = 1, so x = 1/4. Thus C(1/4, 1).</p><p><strong>Step 4:</strong> Check if C is on arc AOB: As we go from O(0,0) to A(4,−4) to B(9,6), the point C(1/4, 1) lies on the arc between O and A on the upper branch. ✓</p><p><strong>Step 5:</strong> Calculate distance from C(1/4, 1) to line 2x − y − 12 = 0: d = |2(1/4) − 1 − 12|/√(4+1) = |1/2 − 1 − 12|/√5 = |−12.5|/√5 = 12.5/√5 = 2.5√5</p><p><strong>Step 6:</strong> Find length AB: AB = √[(9−4)² + (6+4)²] = √[25 + 100] = √125 = 5√5</p><p><strong>Step 7:</strong> Area of △ACB = (1/2) × base × height = (1/2) × 5√5 × 2.5√5 = (1/2) × 5√5 × 2.5√5 = (1/2) × 12.5 × 5 = 31.25</p><p>∴ <strong>Answer: 31.25 sq. units</strong></p>
Correct Answer: 31