问题 解答题
已知椭圆C:
x2
a2
+
y2
b2
=1(a>b>0)过点(1,
3
2
),且长轴长等于4.
(I)求椭圆C的方程;
(II)F1,F2是椭圆C的两个焦点,⊙O是以F1,F2为直径的圆,直线l:y=kx+m与⊙O相切,并与椭圆C交于不同的两点A,B,若
OA
OB
=-
3
2
,求k的值.
答案

(I)有题义长轴长为4,即2a=4,解得:a=2,

∵点(1,

3
2
)在椭圆上,∴
1
4
+
9
4b2
=1
 解得:b2=3

椭圆的方程为:

x2
4
+
y2
3
=1;

(II)由直线l与圆O相切,得:

|m|
1+k2
=1,即:m2=1+k2

设A(x1,y1)B(x2,y2)    由

x2
4
+
y2
3
=1
y=kx+m
  消去y,

整理得:(3+4k2)x2+8kmx+4m2-12=0,

x1+x2=-

8km
3+4k2
x1x2=
4m2-12
3+4k2

∴y1y2=(kx1+m)(kx2+m)=k2x1x2+km(x1+x2)+m2=k2

4m2-12
3+4k2
+km(-
8km
3+4k2
)+m2=
3m2-12k2
3+4k2
x1x2+y1y2=
4m2-12
3+4k2
+
3m2+2k2
3+4k2
=
7m2-12k2-12
3+4k2

∵m2=1+k2x1x2+y1y2=

-5-5k2
3+4k2
=-
3
2

解得:k2=

1
2

k的值为:±

2
2

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