试题与答案

已知椭圆x2a2+y2b2=1(a>b>o)的左焦点为F(-2,0),离心率e=

题型:解答题

题目:

已知椭圆
x2
a2
+
y2
b2
=1(a>b>o)
的左焦点为F(-
2
,0),离心率e=
2
2
,M、N是椭圆上的动点.
(Ⅰ)求椭圆标准方程;
(Ⅱ)设动点P满足:
OP
=
OM
+2
ON
,直线OM与ON的斜率之积为-
1
2
,问:是否存在定点F1,F2,使得|PF1|+|PF2|为定值?,若存在,求出F1,F2的坐标,若不存在,说明理由.
(Ⅲ)若M在第一象限,且点M,N关于原点对称,点M在x轴上的射影为A,连接NA 并延长交椭圆于点B,证明:MN⊥MB.

答案:

(Ⅰ)由题设可知:

c=
2
c
a
=
2
2
,∴a=2,c=
2
…2分

∴b2=a2-c2=2…3分

∴椭圆的标准方程为:

x2
4
+
y2
2
=1…4分

(Ⅱ)设P(xP,yP),M(x1,y1),N(x2,y2),由

OP
=
OM
+2
ON
可得:
xP=x1+2x2
yP=y1+2y2
①…5分

由直线OM与ON的斜率之积为-

1
2
可得:
y1y2
x1x2
=-
1
2
,即x1x2+2y1y2=0②…6分

由①②可得:xP2+2yP2=(x12+2y12)+(x22+2y22

∵M、N是椭圆上的点,∴x12+2y12=4,x22+2y22=4

∴xP2+2yP2=8,即

x2P
8
+
y2P
4
=1…..8分

由椭圆定义可知存在两个定点F1(-2,0),F2(2,0),使得动点P到两定点距离和为定值4

2
;….9分;

(Ⅲ)证明:设M(x1,y1),B(x2,y2),则x1>0,y1>0,x2>0,y2>0,x1≠x2,A(x1,0),N(-x1,-y1)…..10分

由题设可知lAB斜率存在且满足kNA=kNB,∴

y1
2x1
=
y2+y1
x2+x1
….③

kMN•kMB+1=

y1
x1
y2-y1
x2-x1
+1④…12分

将③代入④可得:kMN•kMB+1=

2(y2+y1)
x2+x1
y2-y1
x2-x1
+1=
(
x22
+2
y22
)-(
x21
+2
y21
)
x22
-
x21
⑤….13分

∵点M,B在椭圆

x2
4
+
y2
2
=1上,∴kMN•kMB+1=
(
x22
+2
y22
)-(
x21
+2
y21
)
x22
-
x21
=0

∴kMN•kMB+1=0

∴kMN•kMB=-1

∴MN⊥MB…14分.

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