题目内容

已知抛物线C:y2=4x,动直线l:y=k(x+1)与抛物线C交于A,B两点,O为原点.
(1)求证:
OA
OB
是定值;
(2)求满足
OM
=
OA
+
OB
的点M的轨迹方程.
y2=4x
y=k(x+1)
得k2x2+(2k2-4)x+k2=0.?
由k≠0,且△>0,得-1<k<1,且k≠0.?
设A(x1,y1),B(x2,y2),则x1+x2=
4
k2
-2,x1x2=1.?
(1)证明:
OA
OB
=x1x2+y1y2
=x1x2+k2(x1+1)(x2+1)?
=(k2+1)x1x2+k2(x1+x2)+k2?
=k2+1+k2
4
k2
-2
)+k2=5,?
OA
OB
=5
为常数.?
(2)
OM
=
OA
+
OB
=(x1+x2,y1+y2)=(
4
k2
-2
4
k
).?
设M(x,y),则
x=
4
k2
-2
y=
4
k
消去k得y2=4x+8.?
又∵x=
4
k2
-2
>2,故M的轨迹方程为y2=4x+8(x>2).
练习册系列答案
相关题目

违法和不良信息举报电话:027-86699610 举报邮箱:58377363@163.com

精英家教网