Question

已知向量=(2,0),==(0,1),动点M到定直线y=1的距离等于d,并且满足=K(-d²),其中O为坐标原点,K为参数.

(1)求动点M的轨迹方程,并判断曲线类型;

(2)如果动点M的轨迹是一条圆锥曲线,其离心率e满足≤e≤,求实数K的取值范围.

Answer 1

解：(1)设M(x,y),则由=(2,0),==(0,1),且O为原点得A(2,0),B(2,1),C(0,1).

从而=(x,y),=(x-2,y),=(x,y-1),=(x-2,y-1),

d=|y-1|.                                                                   

代入=K(-d²)得(1-K)x²+2(K-1)x+y²=0为所求轨迹方程.       

当K=1时,得y=0,轨迹为一条直线;                                             

当K≠1时,得(x-1)²+=1.

若K=0,则为圆;                                                            

若K＞1,则为双曲线;                                                        

若0＜K＜1或K＜0,则为椭圆.                                                  

(2)因为≤e≤,所以方程表示椭圆.                                      

对于方程(x-1)²+=1,

①当0＜K＜1时,a²=1,b²=1-K,c²=a²-b²=1-(1-K)=K,

此时e²==K.而≤e≤,所以≤K≤.                              

②当K＜0时,a²=1-K,b²=1,c²=-K,

所以e²=,即≤≤.

所以-1≤K≤.                                                          

所以K∈［-1,］∪［］.