推 zxcv0011 :答案是16沒錯，非常感謝!!! 04/27 09:59
※ 引述《zxcv0011 (Cedric)》之銘言：
: F1,F2為橢圓 x^2/81+y^2/32之兩焦點，AB為過F1之一焦弦，已知三角形ABF2之面積為32,求AB長?
= 1?
a = 9
b = 4√2
c = √[81 - 32] = 7
e = c/a
0 < θ < π
r(θ) = (b^2/a)/[1 - (c/a)cosθ]
r(π+θ) = (b^2/a)/[1 + (c/a)cosθ]
Δ = (1/2)(b^2/a)[2/(1 - (c/a)^2(cosθ)^2)]2csinθ
=> 32 = 14sinθ(32/9)/[1 - (49/81)[1 - (sinθ)^2]]
=> 49(sinθ)^2 - 126(sinθ) + 32 = 0
=> sinθ = 2/7
AB = Δ/[csinθ] = 32/2 = 16
: 先感謝各位大大了
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※ 編輯: Honor1984 (220.141.65.124), 04/27/2014 01:14:28