If AB is the chord passing through the center of the ellipse x2 / 25 + Y2 / 16 = 1 and F1 is the focus of the ellipse, then the area of the triangle f1ab is the largest Method 2 (basic method): s △ f1ab = s △ oaf1 + s △ obf1 = (C × y1-y2) × 2 = (3 × y1-y2) × 2 Case (1): when k exists, let AB: y = KX be substituted into the ellipse x2 / 25 + Y2 / 16 to find y1-y2, and get 60 × √ K square × √ [25 × K square × 16] / (25 × K square × 16) ＜ 12 Case (2): when K does not exist, s = B × 2C △ 2 = 12 According to (1) (2), s ≤ 12 My formula is 60 ×√ K square × √ [25 × K square + 16] / (25 × K square + 16), then I get 15,

If AB is the chord passing through the center of the ellipse x2 / 25 + Y2 / 16 = 1 and F1 is the focus of the ellipse, then the area of the triangle f1ab is the largest Method 2 (basic method): s △ f1ab = s △ oaf1 + s △ obf1 = (C × y1-y2) × 2 = (3 × y1-y2) × 2 Case (1): when k exists, let AB: y = KX be substituted into the ellipse x2 / 25 + Y2 / 16 to find y1-y2, and get 60 × √ K square × √ [25 × K square × 16] / (25 × K square × 16) ＜ 12 Case (2): when K does not exist, s = B × 2C △ 2 = 12 According to (1) (2), s ≤ 12 My formula is 60 ×√ K square × √ [25 × K square + 16] / (25 × K square + 16), then I get 15,

Is it 60 √ (K & # 178;) · √ (25K & # 178; + 16) / (25K & # 178; + 16)?
y=kx,x=y/k,k≠0
x²/16 + y²/25=1
25x²+16y²=400
25(y/k)² + 16y²=400
25y²+16k²y²=400k²
(25+16k²)y²=400k²
y=±√[400k²/(25+16k²)]
|y1-y2|=2√[400k²/(25+16k²)]
=40√[k²/(25+16k²)]
S=(3/2)|y1-y2|
=60√[k²/(25+16k²)]
=60/√[25 + (16/k²)]
25 + 16/k²＞25
√(25 + 16/k²) ＞5
60/√(25 + 16/k²) ＜12
When k = 0, the area is 0