题目内容
已知椭圆
(1)若l与x轴相交于点N,且A是MN的中点,求直线l的方程;
(2)设P为椭圆上一点,且
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【答案】分析:(1)设A(x1,y1),因为A为MN的中点,且M的纵坐标为3,N的纵坐标为0,进而求得yl,又根据点A在椭圆C上,
代入即可求得x1,则点A的坐标可求.
(2)设直线AB的方程和点A,B,P的坐标,把直线方程与椭圆方程联立消去y,根据判别式大于0求得k的范围,根据韦达定理表示出x1+x2和x1x2,进而表示出AB的长度,求得k的范围,进而根据
+
=λ
可知(x1,y1)十(x2,y2)=λ(x3,y3),进而分当λ≠0和λ=0时根据k的范围确定λ的取值范围.
解答:解:(1)设A(x1,y1),
因为A为MN的中点,且M的纵坐标为3,N的纵坐标为0,
所以yl=
,
又因为点A(xl,yl)在椭圆C上,
所以x12+
=1,即
=1,解得x1=±
,
则点A的坐标为(
,
)或(-
,
),
所以直线l的方程为6
x-7y+21=0或6
x+7y-21=0.
(2)设直线AB的方程为y=kx+3或x=0,A(x1,y1),B(x2,y2),P(x3,y3),
当AB的方程为x=0时,|AB|=4>
,与题意不符.
当AB的方程为y=kx+3时:
由题设可得A、B的坐标是方程组
的解,
消去y得(4+k2)x2+6kx+5=0,
所以△=(6k)2-20(4+k2)>0,即k2>5,
则x1+x2=
,x1•x2=
,y1+y2=(kx1+3)+(kx2+3)=
因为|AB|=
,
所以
•
,解得-
<k2<8
所以5<k2<8.
因为
+
=λ
,即(x1,y1)十(x2,y2)=λ(x3,y3),
所以当λ=0时,由
+
=0,得x1+x2=
=0,y1+y2=
=0,
上述方程无解,所以此时符合条件的直线l不存在;
当λ≠0时,x3=
=-
,y3=
因为点P(x3,y3)在椭圆上,
所以[
]2+
[
]2=1化简得λ2=
,
因为5<k2<8,所以3<λ2<4,
则λ∈(-2,-
)∪(
,2).
综上,实数λ的取值范围为(-2,-
)∪(
,2).
点评:本题主要考查了椭圆的简单性质,直线与椭圆的关系,解析几何的知识,解不等式.考查了学生综合分析问题和解决问题的能力.
代入即可求得x1,则点A的坐标可求.
(2)设直线AB的方程和点A,B,P的坐标,把直线方程与椭圆方程联立消去y,根据判别式大于0求得k的范围,根据韦达定理表示出x1+x2和x1x2,进而表示出AB的长度,求得k的范围,进而根据
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解答:解:(1)设A(x1,y1),
因为A为MN的中点,且M的纵坐标为3,N的纵坐标为0,
所以yl=

又因为点A(xl,yl)在椭圆C上,
所以x12+
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则点A的坐标为(
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所以直线l的方程为6
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(2)设直线AB的方程为y=kx+3或x=0,A(x1,y1),B(x2,y2),P(x3,y3),
当AB的方程为x=0时,|AB|=4>

当AB的方程为y=kx+3时:
由题设可得A、B的坐标是方程组

消去y得(4+k2)x2+6kx+5=0,
所以△=(6k)2-20(4+k2)>0,即k2>5,
则x1+x2=

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因为|AB|=
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所以
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所以5<k2<8.
因为
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所以当λ=0时,由
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上述方程无解,所以此时符合条件的直线l不存在;
当λ≠0时,x3=

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因为点P(x3,y3)在椭圆上,
所以[
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因为5<k2<8,所以3<λ2<4,
则λ∈(-2,-
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综上,实数λ的取值范围为(-2,-

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点评:本题主要考查了椭圆的简单性质,直线与椭圆的关系,解析几何的知识,解不等式.考查了学生综合分析问题和解决问题的能力.
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