题目内容
在平面直角坐标系xOy中,已知椭圆E:

(1)求椭圆E的方程;
(2)设P,A,B是椭圆E上异于顶点的三点,Q(m,n)是单位圆x2+y2=1上任一点,使

①求证:直线OA与OB的斜率之积为定值;
②求OA2+OB2的值.
【答案】分析:(1)由△FMN面积
,可得cb=1,再有离心率公式
及a2=b2+c2即可得到a,b,c;
(2)①设P(x,y),A(x1,y1),B(x2,y2),把点A,B代入椭圆方程可得
③,
④,又Q(m,n)是单位圆x2+y2=1上任一点可得m2+n2=1⑤,
由
,得到
因P在椭圆上,故
. 把③④⑤代入上式即可得出x1,y1,x2,y2,满足的式子即可证明结论;
②利用①的结论
为定值.可得故
. 及 又
,可得
.故可证明OA2+OB2=
为定值.
解答:解:(1)由椭圆的离心率为
,得
①,
又△FMN面积
,所以cb=1②,
由①②及a2=b2+c2可解得:a2=2,b2=c2=1,
故椭圆E的方程是
.
(2)①设P(x,y),A(x1,y1),B(x2,y2),
则
③,
④,
又m2+n2=1⑤,
因
,故
因P在椭圆上,故
.
整理得
.
将③④⑤代入上式,并注意点Q(m,n)的任意性,得:
.
所以,
为定值.
②
,
故
.
又
,故
.所以OA2+OB2=
=3.
点评:本题主要考查椭圆的方程与性质、直线方程、直线与椭圆的位置关系、向量运算、斜率的计算公式、三角形的面积计算公式等基础知识,需要较强运算能力、推理论证以及分析问题、解决问题的能力,考查数形结合、化归与转化思想.


(2)①设P(x,y),A(x1,y1),B(x2,y2),把点A,B代入椭圆方程可得


由



②利用①的结论





解答:解:(1)由椭圆的离心率为


又△FMN面积

由①②及a2=b2+c2可解得:a2=2,b2=c2=1,
故椭圆E的方程是

(2)①设P(x,y),A(x1,y1),B(x2,y2),
则


又m2+n2=1⑤,
因


因P在椭圆上,故

整理得

将③④⑤代入上式,并注意点Q(m,n)的任意性,得:

所以,

②

故

又



点评:本题主要考查椭圆的方程与性质、直线方程、直线与椭圆的位置关系、向量运算、斜率的计算公式、三角形的面积计算公式等基础知识,需要较强运算能力、推理论证以及分析问题、解决问题的能力,考查数形结合、化归与转化思想.

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