抛物线y=2x2上两点A(x1,y1),B(x2,y2)关于直线y=x+m对称,若2x1x2=-1,则2m的值是( ) A.3 B.4 C.5 D.6
问题描述:
抛物线y=2x2上两点A(x1,y1),B(x2,y2)关于直线y=x+m对称,若2x1x2=-1,则2m的值是( )
A. 3
B. 4
C. 5
D. 6
答
由
=−1,
y1−y2
x1−x2
=
y1+y2
2
+m,2x1x2=−1,以及y1=2x12,y2=2x22 可得
x1+x2
2
x2−x1=y1−y2=2(
−
x
21
),⇒x1+x2=−
x
22
,1 2
2m=(y1+y2)−(x1+x2)=2(
+
x
21
)+
x
22
=2(x1+x2)2−4x1x2+1 2
=2•1 2
+2+1 4
=3,1 2
故选 A.