【轉(zhuǎn)】三河三中高二20xx年寒假作業(yè)(數(shù)學(xué))答案【理科】-資料下載頁(yè)
2025-08-11 12:16本頁(yè)面
【正文】 2/(x2 + 1)] = 1 ，即(y1y2)/[(x1x2) + (x1 + x2) + 1] = 1......T式將y = kx + 2代入橢圓方程得：x^2 + 3(kx + 2)^2 = 3或3y^2 + [(y 2)^2/(k^2)] = 3分別整理得：(1 + 3k^2)x^2 + 12kx + 9 = 0以及：[3 + (1/k^2)]y^2 (4y/k^2) + [(4/k^2) 3] = 0根據(jù)韋達(dá)定理：x1x2 = 9/(1 + 3k^2) ，x1 + x2 = 12k/(1 + 3k^2)，y1y2 = (4 3k^2)/(1 + 3k^2)∴(x1x2) + (x1 + x2) + 1 = (10 12k + 3k^2)/(1 + 3k^2)，代入T式：(4 3k^2)/(10 12k + 3k^2) = 1∴6k^2 12k + 6 = 0，解得k = 1，直線為：y = x + 2因此，存在k = 1使得 以CD為直徑的圓過(guò)點(diǎn)E
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