已知A(4,0),N(1,0),若点P满足
(1)求点P的轨迹方程,并说明该轨迹是什么曲线; (2)求|
(3)若M(-1,0),求∠MPN在[0,π]上的取值范围. |
(1)设P(x,y),
=(x-4,y),AP
=(1-x,-y),PN
=(-3,0),AN
∵
•AN
=6||,AP
∴-3(x-4)=6
,即3x2+4y2=12.(1-x)2+(-y)2
∴
+x2 4
=1.∴P点的轨迹是以(-1,0)、(1,0)为焦点,长轴长为4的椭圆.y2 3
(2)N(1,0)为椭圆的右焦点,x=4为右准线,
设P(x0,y0),P到右准线的距离为d,d=4-x0,
=e=|PN| d
,|PN|=1 2
d=1 2
.4-x0 2
∵-2≤x0≤2,∴1≤|PN|≤3.
当|PN|=1时,P(2,0);当|PN|=3时,P(-2,0).
(3)令|PN|=t(1≤t≤3),
则|PM|=4-t,|MN|=2,
cos∠MPN=
=|PN|2+|PM|2-|MN|2 2|PN||PM|
=-1+t2+(4-t)2-4 2t(4-t)
.6 t(4-t)
由1≤t≤3,得3≤t(4-t)≤4,
∴
≤cos∠MPN≤1,1 2
∴0≤∠MPN≤
.π 3