设P(x,y)则MP=(x+1,y);NP=(x-1,y);PM=(-x-1,-y);PN=(1-x,-Y)
NM=(-2,0);MN=(2,0)
MP*MN=2X+2PM*PN=x~2-1+y~2NM*NP=2-2x
由已知有2x+2+2-2x=(X~2-1+y~2)*2即x~2+y~2=3
故P为以(0,0)为圆心根号3为半径的圆
2。 PM=(-1-X0,-Y0);PN=(1-X0,-Y0);
则COS(SITA)=(X0~2-1+Y0~2)/根号{[(1+X0)~2+Y0~2]*(1-X0)~2+YO~2}
=
对“点P的轨迹是什么曲线”其他回答
设P(X,Y)
MP=(X+1,Y) NP=(X-1,Y) MN=(2,0)
MP*MN=2X+2PM*PN=X*X+Y*Y-1 NM*NP=2X-2
所以(2X+2)+(2X-2)=4X=2*(X*X=Y*Y-1)
整理得(X-1)*(X-1)-Y*Y=2

已解决