孤射雕
幼苗
共回答了19个问题采纳率:84.2% 举报
∵e=
c
a
=
3
2
,∴设椭圆方程为
x2
4b2
+
y2
b2
=1,
将M(2,1)代入,得
4
4b2
+
1
b2
=1,解得b2=2,
所以椭圆C的方程为
x2
8
+
y2
2
=1,
因此左焦点为(-
6
,0),斜率k1=kOM=
1
2
,
所以直线l的方程为y=
1
2
(x+
6
),即y=
1
2
x+
6
2
.
(2)证明:设直线MA,MB的斜率分别为k1,k2,A(x1,y1),B(x2,y2),
则k1=
y1−1
x1−2
,k2=
y2−1
x2−2
,
∴k1+k2=
y1−1
x1−2
+
y2−1
x2−2
=
(y1−1)(x2−2)+(y2−1)(x1−2)
(x1−2)(x2−2)
=
(
1
2
x1+m−1)(x2−2)+(
1
2
x2+m−1)(x1−2)
(x1−2)(x2−2)
=
x1x2+(m−2)(x1+x2)−4(m−1)
(x1−2)(x2−2)
,(*)
设l:y=
1
2
x+m,由
y=
1
2
x+m
x2
8
+
y2
2
=1
,得x2+2mx+2m2-4=0,
所以x1+x2=-2m,x1x2=2m2−4,
代入(*)式,得
k1+k2=
2m2−4+(m−2)(−2m)−4(m−1)
(x1−2)(x2−2)
=
2m2−4−2m2+4m−4m+4
(x1−2)(x2−2)
=0.
所以直线MA,MB与x轴总围成等腰三角形.
1年前
6