<p>Let P be a variable point on the ellipse \(\frac{x^2}{25} + \frac{y^2}{4} = 2\) with foci \(F_1\) and \(F_2\). If A is the area of the \(\triangle PF_1F_2\) then the maximum value of A is ___.</p>
Step-by-Step Solution
Key Concept: Rewrite the ellipse in standard form to identify a and b, then use the focal chord property: area of triangle with foci is maximized when P is at the minor axis endpoints where height is maximum.
<p><strong>Step 1: Convert to standard form</strong></p><p>Divide the equation by 2: $\frac{x^2}{50} + \frac{y^2}{8} = 1$</p><p>Here a² = 50, b² = 8, so a = 5√2, b = 2√2</p><p><strong>Step 2: Find the distance between foci</strong></p><p>c² = a² - b² = 50 - 8 = 42</p><p>c = √42</p><p>Distance F₁F₂ = 2c = 2√42</p><p><strong>Step 3: Maximize the area of △PF₁F₂</strong></p><p>Area = ½ × base × height = ½ × 2c × h, where h is the perpendicular distance from P to the major axis</p><p>Maximum height occurs when P is at the endpoints of the minor axis, so h_max = b = 2√2</p><p><strong>Step 4: Calculate maximum area</strong></p><p>A_max = ½ × 2√42 × 2√2 = √42 × 2√2 = 2√84 = 2 × 2√21 = 4√21</p><p>∴ Answer: <strong>4√21</strong></p>
Correct Answer: 4