(Ⅰ) k DE∈[-1,1].
(Ⅱ) 所求轨迹方程为: x 2=4 y , x ∈[-2,2]
解法一: 如图, (Ⅰ)设 D ( x 0, y 0), E ( x E, y E), M ( x , y ).由= t ,
= t , 知( x D-2, y D-1)= t (-2,-2).
∴ 同理 .
∴ k DE= = = 1-2 t .
∵
t ∈[0,1] , ∴ k DE∈[-1,1].
(Ⅱ) ∵= t ∴( x +2 t -2, y +2 t -1)= t (-2 t +2 t -2,2 t -1+2 t -1)
= t (-2,4 t -2)=(-
2 t ,4 t 2-2 t ).
∴ , ∴ y =, 即 x 2=4 y .∵ t ∈[0,1], x =2(1-2 t )∈[-2,2].
即所求轨迹方程为: x 2=4 y , x ∈[-2,2]
解法二:
(Ⅰ)同上.
(Ⅱ) 如图, ="+" =" + " t =" +" t (-) = (1- t ) + t ,
=" +" = + t = + t (-) =(1- t ) + t ,
=" +=" + t = + t (-)=(1- t ) + t
= (1- t 2) + 2(1- t ) t + t 2.
设 M 点的坐标为( x , y ),由=(2,1), =(0,-1), =(
-2,1)得
消去 t 得 x 2=4 y
∵ t ∈[0,1], x ∈[-2,2].
故所求轨迹方程为: x 2=4 y , x ∈[-2,2]