问题
解答题
已知椭圆C:
(Ⅰ)求椭圆C的方程; (Ⅱ)已知动直线y=k(x+1)与椭圆C相交于A、B两点. ①若线段AB中点的横坐标为-
②若点M(-
|
答案
(Ⅰ)因为
+x2 a2
=1(a>b>0)满足a2=b2+c2①,y2 b2
由
=c a
②,2b=6 3
③.联立①②③,2 15 3
解得a2=5,b2=
,5 3
所以椭圆方程为
+x2 5
=1.y2 5 3
(Ⅱ)(1)将y=k(x+1)代入
+x2 5
=1中,得(1+3k2)x2+6k2x+3k2-5=0,y2 5 3
△=36k4-4(3k2+1)(3k2-5)=48k2+20>0,x1+x2=-
,6k2 3k2+1
因为AB中点的横坐标为-
,所以-1 2
=-6k2 3k2+1
,解得k=±1 2
,3 3
(2)由(1)知x1+x2=-
,x1x2=6k2 3k2+1
,3k2-5 3k2+1
所以
•MA
=(x1+MB
,y1)(x2+7 3
,y2)=(x1+7 3
)(x2+7 3
)+y1y27 3
=(x1+
)(x2+7 3
)+k2(x1+1)(x2+1)7 3
=(1+k2)x1x2+(
+k2)(x1+x2)+7 3
+k249 9
=(1+k2)
+(3k2-5 3k2+1
+k2)(-7 3
)+6k2 3k2+1
+k2=49 9
;4 9