题干

已知椭圆 C
x
2
a
2
+
y
2
b
2
=1(a>b>0)
的左,右焦点分别为F1,F2,过F1任作一条与两坐标轴都不垂直的直线,与C交于A,B两点,且△ABF2的周长为8.当直线AB的斜率为
3
4
时,AF2与x轴垂直.

(I)求椭圆C的方程;

(Ⅱ)在x轴上是否存在定点M,总能使MF1平分∠AMB?说明理由.

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答案(点此获取答案解析)

解:(I)由椭圆的定义可知△ABF2的周长4a=8,则a=2,
由直线AB的斜率为
3
4
时,AF2与x轴垂直,则tan∠AF1F2=

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