狼本
幼苗
共回答了20个问题采纳率:80% 举报
(I)抛物线x2=4
2
y的焦点坐标为(0,
2
),可得椭圆的上顶点为(0,
2
),得b=
2
∵椭圆的离心率e=
3
3
,得
c
a
=
3
3
,解得a=
3
,c=1
∴椭圆C的方程是
x2
3
+
y2
2
=1
(II)由(I)得椭圆C的右焦点为F2(1,0)
①当直线l与x轴垂直时,直线l斜率不存在,此时M(1,
2
3
3
),N(1,-
2
3
3
)
∴
OM
•
ON
=1×1+
2
3
3
×(-
2
3
3
)=-
1
3
,不符合题意;
②当直线l与x轴不垂直时,设直线方程l:y=k(x-1),且M(x1,y1),N(x2,y2)
由
x2
3
+
y2
2
=1
y=k(x−1)
,得(2+3k2)x2-6k2x+3k2-6=0
x1+x2=
6k2
2+3k 2
,x1•x2=
3k2−6
2+3k2
∴
OM
•
ON
=x1•x2+y1•y2=x1•x2+k2[x1•x2-(x1+x2)+1]=(1+k2)x1•x2-k2(x1+x2)+k2=-1
即(1+k2)•
3k2−6
2+3k2
-k2•
6k2
2+3k 2
+k2=-1
解之得k=±
2
,故直线l的方程是y=
2
(x-1)或y=-
2
(x-1).
(III)设M(x1,y1),N(x2,y2),A(x3,y3),B(x4,y4)
由(II)得|MN|=
(x1−x2)2+(y1−y2)2
=
1+k2
|x1-x2|
=
(1+k2)[(x1+x2)2−4x 1x2]
=
(1+k2)[(
6k2
2+3k 2
)2−4×
3k2−6
2+3k2
]
=
4
3
(k2+1)
2+3k2
由
x2
3
+
y2
2
=1
y=kx
消去y,整理得x2=
6
2+3k 2
∴|AB|=
(x3−x4)2+(y3−y4)2
=
1+k2
|x3-x4|=2
6(k2+1)
2+3k2
∴
3
|AB|2
|MN|
=
24
3
(k2+1)
2+3k2
4
3
(k2+1)
2+3k2
=6.
1年前
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