在平面直角坐标系xOy中,已知椭圆 C: x 2 a 2 + y 2 b 2 =1(a>

2025-05-22 05:02:22
推荐回答(1个)
回答1:

(1)由题意:
c=1
e=
c
a
=
1
2
a 2 = b 2 + c 2
,解得: a=2,b=
3

所以椭圆C:
x 2
4
+
y 2
3
=1

(2)由(1)可知 A 1 (0,
3
), A 2 (0,-
3
)
,设Q(x 0 ,y 0 ),
直线QA 1 y-
3
=
y 0 -
3
x 0
x
,令y=0,得 x S =
-
3
x 0
y 0 -
3
;     
直线QA 2 y+
3
=
y 0 +
3
x 0
x
,令y=0,得 x T =
3
x 0
y 0 +
3

|OS|?|OT|=|
-
3
x 0
y 0 -
3
?
3
x 0
y 0 +
3
|=|
3
x 20
y 20
-3
|

x 20
4
+
y 20
3
=1
,所以 3
x 20
=4(3-
y 20
)

所以 |OM|?|ON|=|
3
x 20
y 20
-3
|=4

(3)假设存在点M(m,n)满足题意,则
m 2
4
+