1设P(x,y)由:向量AP 点乘 向量BP = K,K=(向量PC)2得:(x,y-1)(x,y+1)=(1-x,-y)2x2+y2-1=(1-x)2+y2化简得:x=1,即动点P的方程。动点P是直线(过(1,0)点,平行于y轴)2.K=2时:x2+y2-1=2x2+y2=3此时,动点P轨迹是圆,圆心在坐标原点,半径√3。同时可得到x,y此时的定义域,都是(-√3,√3)。目标函数f=|2(向量AP)+向量BP|=|2(x,y-1)+(x,y+1)|=|(3x,3y-1)|=√(9x2+9y2+1-6y)=√(27+1-6y)
…… …… x2+y2=3代入=√(28-6y)所以此函数f的单调性仅跟变量y相关,是关于y的减函数。当y=√3时,f有最小值,f=√(28-6√3)当y=-√3时,f有最大值,f=√(28+6√3) 2令t=(x+y-1)/(x-y+3)则整理得:x=[(t+1)/(t-1)](y-1)+2t/(1-t) ①由x^2+(y-1)^2≤4,则一定有(y-1)^2≤4-x^2≤4;则-4≤y-1≤4;令ξ=y-1;即-4≤ξ≤4①代入x^2+(y-1)^2≤4得:{[(t+1)/(t-1)]^2+1}·ξ^2 -[2t/(t-1)^2]·ξ+4t^2/(t-1)^2≤4.即{[(t+1)/(t-1)]^2+1}·ξ^2 -[2t/(t-1)^2]·ξ+4(2t-1)/(t-1)^2≤0而这个关于ξ的不等式的解为:-4≤ξ≤4取当ξ=-4和ξ=4时式子{[(t+1)/(t-1)]^2+1}·ξ^2 -[2t/(t-1)^2]·ξ+4(2t-1)/(t-1)^2的值,使之分别≤0,则可求出t的最值;亦即(x+y-1)/(x-y+3)的最值. |