题目内容
(本小题满分12分)
如图,三定点A(2,1),B(0,-1),C(-2,1); 三 动点D,E,M满足="t," =" t" ,
="t" , t∈[0,1].
(Ⅰ) 求动直线DE斜率的变化范围;
(Ⅱ) 求动点M的轨迹方程.
如图,三定点A(2,1),B(0,-1),C(-2,1); 三 动点D,E,M满足="t," =" t" ,
="t" , t∈[0,1].
(Ⅰ) 求动直线DE斜率的变化范围;
(Ⅱ) 求动点M的轨迹方程.
(Ⅰ) kDE∈[-1,1].
(Ⅱ) 所求轨迹方程为: x2=4y, x∈[-2,2]
(Ⅱ) 所求轨迹方程为: x2=4y, x∈[-2,2]
解法一: 如图, (Ⅰ)设D(x0,y0),E(xE,yE),M(x,y).由=t,
= t, 知(xD-2,yD-1)=t(-2,-2).
∴ 同理 .
∴kDE = = = 1-2t.
∵t∈[0,1] , ∴kDE∈[-1,1].
(Ⅱ) ∵=t ∴(x+2t-2,y+2t-1)=t(-2t+2t-2,2t-1+2t-1)
=t(-2,4t-2)=(-2t,4t2-2t).
∴ , ∴y=, 即x2=4y.∵t∈[0,1], x=2(1-2t)∈[-2,2].
即所求轨迹方程为: x2=4y, 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-t2) + 2(1-t)t+t2.
设M点的坐标为(x,y),由=(2,1), =(0,-1), =(-2,1)得
消去t得x2=4y
∵t∈[0,1], x∈[-2,2].
故所求轨迹方程为: x2=4y, x∈[-2,2]
= t, 知(xD-2,yD-1)=t(-2,-2).
∴ 同理 .
∴kDE = = = 1-2t.
∵t∈[0,1] , ∴kDE∈[-1,1].
(Ⅱ) ∵=t ∴(x+2t-2,y+2t-1)=t(-2t+2t-2,2t-1+2t-1)
=t(-2,4t-2)=(-2t,4t2-2t).
∴ , ∴y=, 即x2=4y.∵t∈[0,1], x=2(1-2t)∈[-2,2].
即所求轨迹方程为: x2=4y, 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-t2) + 2(1-t)t+t2.
设M点的坐标为(x,y),由=(2,1), =(0,-1), =(-2,1)得
消去t得x2=4y
∵t∈[0,1], x∈[-2,2].
故所求轨迹方程为: x2=4y, x∈[-2,2]
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