题目内容

已知直线l:y=x+m与抛物线y2=8x交于A、B两点,
(1)若|AB|=10,求m的值;
(2)若OA⊥OB,求m的值.
设A(x1,y1)、B(x2,y2
(1)
y=x+m
y2=8x
x2+(2m-8)x+m2=0------------------------------(1分)
△=(2m-8)2-4m2>0
x1+x2=8-2m
x1x2=m2
-----------------------------------------------(3分)|AB|=
2
|x1-x2|=
2
(x1+x2)2-4x1x2
=10
m=
7
16
----(5分)
∵m<2,∴m=
7
16
---------------------------------------------------------(6分)
(2)∵OA⊥OB,∴x1x2+y1y2=0------------------------------------(7分)
x1x2+(x1+m)(x2+m)=0,2x1x2+m(x1+x2)+m2=0-----------------------------------------(9分)
2m2+m(8-2m)+m2=0,m2+8m=0,m=0orm=-8,---------------------------------(11分)
经检验m=-8------------------------------------------------------------(12分)
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