Parabola
Area of Triangle with points on Parabola
Grade 11
Question:
<p>Given parabola \(y^2 = 4x\) has two points \(A(4, -4)\) and \(B(9, 6)\). A point \(C(t^2, 2t)\) lies on the parabola. Find the maximum area of triangle \(ABC\) (in sq. units).</p>
Step-by-Step Solution
Key Concept: Parameterize point C on the parabola as (t², 2t), then express triangle area using the determinant formula as |t² - 5t - 12|/2. Maximize this absolute value by finding critical points and boundary behavior.
<p><strong>Step 1: Verify C is on parabola</strong></p><p>Point C(t², 2t) satisfies y² = (2t)² = 4t² = 4(t²) ✓</p><p><strong>Step 2: Calculate triangle area using determinant formula</strong></p><p>Area = ½|x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)|</p><p>= ½|4(6 - 2t) + 9(2t - (-4)) + t²((-4) - 6)|</p><p>= ½|24 - 8t + 18t + 36 - 10t²|</p><p>= ½|-10t² + 10t + 60|</p><p>= 5|−t² + t + 6|</p><p><strong>Step 3: Maximize |−t² + t + 6|</strong></p><p>Let f(t) = −t² + t + 6. This is a downward parabola with vertex at t = −1/(2(−1)) = ½</p><p>f(½) = −¼ + ½ + 6 = 6.25</p><p><strong>Step 4: Check boundary values</strong></p><p>As t → ±∞, f(t) → −∞, so |f(t)| → ∞. But we need the actual maximum of the area function:</p><p>The maximum value of |−t² + t + 6| = 6.25 (at vertex)</p><p>∴ Maximum Area = 5 × 6.25 = <strong>31.25 sq. units</strong></p>
Correct Answer: 31.25