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y1−y2x1−x2=−1,y1+y22=x1+x22+m,2x1x2=−1,以及y1=2x12,y2=2x22 可得 x2−x1=y1−y2=2(x21−x22),⇒x1+x2=−12,2m=(y1+y2)−(x1+x2)=2(x21+x22)+12=2(x1+x2)2−4x1x2+12=2•14+2+12=3,故...