Question

已知向量=(2,0),==(0,1),動點M到定直線y=1的距離等于d,并且滿足=K(-d²),其中O為坐標原點,K為參數.

(1)求動點M的軌跡方程,并判斷曲線類型;

(2)如果動點M的軌跡是一條圓錐曲線,其離心率e滿足≤e≤,求實數K的取值范圍.

Answer 1

解：(1)設M(x,y),則由=(2,0),==(0,1),且O為原點得A(2,0),B(2,1),C(0,1).

從而=(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為所求軌跡方程.       

當K=1時,得y=0,軌跡為一條直線;                                             

當K≠1時,得(x-1)²+=1.

若K=0,則為圓;                                                            

若K＞1,則為雙曲線;                                                        

若0＜K＜1或K＜0,則為橢圓.                                                  

(2)因為≤e≤,所以方程表示橢圓.                                      

對于方程(x-1)²+=1,

①當0＜K＜1時,a²=1,b²=1-K,c²=a²-b²=1-(1-K)=K,

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

②當K＜0時,a²=1-K,b²=1,c²=-K,

所以e²=,即≤≤.

所以-1≤K≤.                                                          

所以K∈［-1,］∪［］.