内容正文:
微专题43 非对称韦达定理
[考情分析] 圆锥曲线的综合问题是高考考查的重点内容,非对称韦达定理的应用在高考中经常出现,常以解答题的形式压轴出现,难度较大.
典例1 设椭圆C:+=1(a>b>0)的左焦点为F,过点F的直线l与椭圆C相交于A,B两点,直线l的倾斜角为60°,=2,求椭圆C的离心率.
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典例2 已知抛物线C:y2=4x与定点P(2,1),直线l过点P且与抛物线交于A,B两点,且有=,求直线l的斜率.
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典例3 如图,设F为椭圆C:+y2=1的右焦点,过点(2,0)的直线l与椭圆C交于A,B两点.
(1)若点B为椭圆C的上顶点,求直线AF的方程;
(2)设直线AF,BF的斜率分别为k1,k2(k2≠0),求证:为定值.
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[总结提升]
在解决直线与圆锥曲线的位置关系的问题中,在遇到直线与圆锥曲线联立转化为一元二次方程,得到韦达定理后,发现不能直接应用韦达定理,这类问题叫做“非对称韦达问题”,处理这类问题常用两种方法,一是和积转换法,二是配凑半代换法.
1.已知圆C:(x+1)2+y2=8,定点A(1,0),M为圆上动点,点P在AM上,点N在CM上,且满足=2,·=0,点N的轨迹为曲线E.
(1)求曲线E的方程;
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(2)过定点F(0,2)的直线交曲线E于不