椭圆的离心率为, 过点, 记椭圆的左顶点为. (1)求椭圆的方程;

 椭圆的离心率为, 过点, 记椭圆的左顶点为.

1)求椭圆的方程;

2)设垂直于轴的直线交椭圆于两点, 试求面积的最大值;

3)过点作两条斜率分别为的直线交椭圆于两点,且, 求证: 直线恒过一个定点.

答案

1)由,解得

所以椭圆C的方程为x22y21.

2) 解:设B(mn)C(mn),则SABC×2|m|×|n||m|·|n|

1m22n2≥22|m|·|n|,所以|m|·|n|≤

当且仅当|m||n|时取等号,

从而SABC,即ABC面积的最大值为.

3)证明:因为A(1,0),所以AByk1(x1)ACyk2(x1)

消去y,得(12k)x24kx2k10,解得x=-1

,同理,有,而k1k22

直线BC的方程为

,即

所以,得直线BC恒过定点.

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