已知抛物线y=-(x^2)+bx+c与x轴交于A、B,与y轴交于点C(0,3),抛物线的顶点为M 对称轴是直线x=1,

1. 与y轴交于点C(0,3), c = 3; 对称轴是直线x=1, -b/[2(-1)] = 1, b = 2抛物线: y = -x² + 2x + 3 = -(x-3)(x+1)A(-1, 0), B(3, 0)设H(h, -h² + 2h + 3), G(0, g)BG² = GH² 9+g² = h² + (h² - 2h - 3 +g)²(1)BG的斜率k = (g - 0)/(0 - 3) = -g/3GH的斜率l = (g + h² -2h -3)/(0-h) = -(h² - 2h - 3 +g)/hBG与GH垂直, kl = -1h² - 2h - 3 +g = -3h/g (2)(2)代入(1): 9+g² = h² + (-3h/g)² = h²(9+g²)/g²h² = g²h = ±g(i) h = g代入(2): h² - h = 0h = 0, H(0, 3), G(0, 3), 舍去h = 1, g = 1H(1, 4)(ii) h = -g代入(2): h² - 3h - 6 = 0h = (3±√33)/2H((3+√33)/2, -2 - 2√33/9)或H((3-√33)/2, -2 +2√33/9)另外还应当考虑BH与BG垂直, 但似乎做起来太麻烦.(2)题中未说清楚,假定为面积最小M(1, 4) MC的方程: (y-3)/(x-0) = (4-3)/(1-0)x - y + 3 = 0D(-3, 0)四边形STDC由三角形SDT和三角形CDS组成三角形SDT的底ST为1, 高为D与对称轴的距离(=4), 面积一定.三角形CDS的底CD为定值, 高为S与CD的距离.当S与M重合时, 高为0, 此时四边形STDC的面积最小(实际为三角形SDT) 如为周长,因为CD和ST均为定值,只需SC和DT之和最小即可.CD=√(3²+3²) = 3√2ST = 1设T(1, t), S(1, t+1)SC = √[(1-0)² + (t+1-2)²] = √[1 + (t-2)²] CT = √[(1+3)² + (t-0)²] = √(16 + t²)w = SC + DT = √[1 + (t-2)²] + √(16 + t²)求导（不知是否有更简单的办法): w' = (t-2)/√[1 + (t-2)²]+ t/√(16 + t²) = 0(t-2)/√[1 + (t-2)²]= -t/√(16 + t²)两边平方： (t-2)²/[1 + (t-2)²] = t²/(16 + t²)(t-2)² = t²/16t-2 = ±t/4t1 = 8/3, w = (√13 + √208)/3≈6.009t2 = 8/5, w = (√29 +√464)/5 ≈ 5.385因平方可能引起增根,验算可知t= 8/5时,w最小(5.385; t=8/3时,w = 6.009)最小周长=CD+ST+w = 3√2 + 1 + (√29 +√464)/53. P(p, -p²+2p+3)P与CD的距离为圆半径4|p -(-p²+2p+3)+3|/√2 = 4|p²-p|/√2 = 4p²-p +4√2 = 0 (此方程无解)p²-p -4√2 = 0p = [1±√(1+16√2)]/2P([1+√(1+16√2)]/2, 0.275)或P(([1-√(1+16√2)]/2, -4.585)（可以自己表达为根号的形式）(4)显然Q在AB的中垂线即抛物线的对称轴上,即Q的横坐标为1.又Q在AC的中垂线上（斜率-1/3,过(-1/2, 3/2)), 其方程为y = (4 - x)/3Q(1, 1)圆Q的半径R=√5要使P圆Q和CD均相切 有两种情形：(1)外切P(1, p)与CD的距离为其半径r = |1-p+3|/√2 = |p - 4|/√2PQ = r+R PQ= |p-1| |p - 4|/√2 +√5 = |p-1|(i) p ≥ 4去掉绝对值号,p = (√10-√2-4)/(√2-1) < 0, 舍去(ii) 1 ≤ p < 4去掉绝对值号,p = (√10+√2+4)/(√2+1), 不与前提冲突.(iii) p < 1 去掉绝对值号,p = (-√10+√2-4)/(√2-1), 不与前提冲突.(2)内切和外切类似地做.p ≥ 4时,解与外切的1 ≤ p < 4情形相同.p < 1时,解为p = (-√10+√2-4)/(√2-1),　不与前提冲突.真是累死了. 部分答案可见插图,没时间全画了.