x2 |
a2 |
y2 |
b2 |
由已知b=
3 |
c |
a |
1 |
2 |
所以a=2, b=
3 |
得橢圓的方程為
x2 |
4 |
y2 |
3 |
(Ⅱ)設(shè)P(x,y),
又A(-2,0),F(xiàn)(1,0),則
PA |
PF |
∴
PA |
PF |
=x2+x-2+y2=
1 |
4 |
當(dāng)x=0時(shí),取得最小值0,當(dāng)x=2時(shí),取得最大值4,
∴
PA |
PF |
x2 |
a2 |
y2 |
b2 |
3 |
1 |
2 |
AP |
FP |
x2 |
a2 |
y2 |
b2 |
3 |
c |
a |
1 |
2 |
3 |
x2 |
4 |
y2 |
3 |
PA |
PF |
PA |
PF |
1 |
4 |
PA |
PF |