請教一下第8題，又畫圖的話A應該不行吧？
若第7題考計算又該如何化簡求值
感謝囉
98全國聯招
版主: thepiano
Re: 98全國聯招
該圓除了與拋物線交於兩點外，它與拋物線亦相切於原點，故應算有相切！
第 7 題
所求 = 12(1 - 1/2^2)(1 - 1/3^2)......(1 - 1/n^2) ( n → ∞ )
(1 - 1/2^2)(1 - 1/3^2)......(1 - 1/n^2) = (1/2) *(3/2) * (2/3) *(4/3) * (3/4) *(5/4) * ...... * [(n - 1)/n] *[(n + 1) / n] = (1/2) * [(n + 1) / n]
當 n → ∞ 時，(1 - 1/2^2)(1 - 1/3^2)......(1 - 1/n^2) = 1/2
第 7 題
所求 = 12(1 - 1/2^2)(1 - 1/3^2)......(1 - 1/n^2) ( n → ∞ )
(1 - 1/2^2)(1 - 1/3^2)......(1 - 1/n^2) = (1/2) *(3/2) * (2/3) *(4/3) * (3/4) *(5/4) * ...... * [(n - 1)/n] *[(n + 1) / n] = (1/2) * [(n + 1) / n]
當 n → ∞ 時，(1 - 1/2^2)(1 - 1/3^2)......(1 - 1/n^2) = 1/2
Re: 98全國聯招
設圓心為(x, 3)，且圓與拋物線的切點為(t, t^2) => 切線斜率為2t，圓半徑為3
則 [(t^2 - 3)/(t - x)]*2t=-1， (t - x)^2 + (t^2 - 3)^2 = 9
=> 4t^6 - 23t^4 + 30t^2 = 0
=> t=0, sqrt(2), -sqrt(2), sqrt(15)/2, -sqrt(15)/2
=> x = 0, sqrt(2), -sqrt(2), 5sqrt(15)/4, -5sqrt(15)/4
則 [(t^2 - 3)/(t - x)]*2t=-1， (t - x)^2 + (t^2 - 3)^2 = 9
=> 4t^6 - 23t^4 + 30t^2 = 0
=> t=0, sqrt(2), -sqrt(2), sqrt(15)/2, -sqrt(15)/2
=> x = 0, sqrt(2), -sqrt(2), 5sqrt(15)/4, -5sqrt(15)/4