举报
ccf21970
因为开始漏了x²的系数(-2), 现在重做. (1) OC = AB = √[(-2 - 0)² + (0 - 2)²] = 2√2 C(0, 2√2) n = 2√2 y = -2x² + mx + 2√2 过A(-2, 0): -8 - 2m + 2√2 = 0 m = √2 - 4 y = -2x² + (√2 - 4)x + 2√2 (2) 不变 (3) AB的方程: x/(-2) + y/2 = 1, y = 2 + x △EOF为等腰三角形时, 显然∠EOF不可能为90˚, 有两种可能:EF = EO或FE = FO (i) EF = EO ∠OEF的平分线与OF垂直,而且平分OF, 此时平分线与x轴平行, OE, EF的斜率显然互为相反数 设E(e, 2 + e), 则F(0, 4 + 2e), -2 < e < 0 EF的斜率p = tan(45°/2) = sin45°/(1 + cos45°) = √2 - 1 OE的斜率q = -p = 1 - √2 OE的方程: y = (1 - √2)x E(-√2, 2 - √2) (ii)FE = FO ∠OEF = ∠FOE = 45°, EF与OF垂直 OE的斜率 = tan(90° + 45°) = -1 OE的方程: y = -x 与AB的交点为E(-1, 1) (4) (3)中的(ii)的结果是EF与x轴平行, 这里不考虑. E(-√2, 2 - √2), F(0, 4 - 2√2) EF的方程: y = (√2 - 1)x + 4 - 2√2, (√2 - 1)x - y + 4 - 2√2 = 0 y = 0, x = -2√2, D(-2√2, 0) EF = √[(-√2 - 0)² + (2 - √2 - 4 + 2√2)²] = √(8 - 4√2) P(p, -2p² + (√2 - 4)p + 2√2) P与EF的距离h = |(√2 - 1)p + 2p² - (√2 - 4)p - 2√2 + 4 - 2√2}/√[(√2 - 1)² + 1²] = |2p² + 3p + 4 - 4√2|/√(4 - 2√2) △EPF的面积S1 = (1/2)*EF*h = (√2/2)|2p² + 3p + 4 - 4√2| 因为P在直线EF的上方, S1 = -(√2/2)(2p² + 3p + 4 - 4√2) (a) PE的方程: ( y - 2 + √2)/[ -2p² + (√2 - 4)p + 2√2 - 2 + √2] = (x + √2)/(p + √2) 取y = 0, x = -√2 + [(√2 - 2)p - 2√2 + 2]/[ -2p² + (√2 - 4)p + 3√2 - 2] G( -√2 + [(√2 - 2)p - 2√2 + 2]/[ -2p² + (√2 - 4)p + 3√2 - 2], 0) △EDG的面积S2 = (1/2)DG*E的纵坐标 = (1/2)(2 - √2){ -√2 + [(√2 - 2)p - 2√2 + 2]/[ -2p² + (√2 - 4)p + 3√2 - 2] + 2√2] (b) S1 = (2√2 + 1)S2 有两个解: p = 0, P(0, 2√2), 与C重合 p = (√2 - 4)/2, P((√2 - 4))/2, 1+ 2√2) 前者很容易看出。后者我是通过图的帮助得到的,应当还有办法,不过我现在没找到。