1抛物线y2=2x上两点A(x1,y1),B(x2,y2)关于直线y=x+b对称,且y1y2=-1,求b=?若x1满足2x+2的x次方=5,x2满足2x+2log2为底(x+1)=5,则x1+x2=?

问题描述:

1抛物线y2=2x上两点A(x1,y1),B(x2,y2)关于直线y=x+b对称,且y1y2=-1,求b=?
若x1满足2x+2的x次方=5,x2满足2x+2log2为底(x+1)=5,则x1+x2=?

由AB两点斜率为-1可得Y1-Y2=X2-X1.(*)y2=2x,可消去(*)式x,整理得Y1+Y2=-2.AB中点在直线上,有:Y1+Y2=X1+X2+2b.结合抛物线有:X1+X2=[(Y1+Y2)平方-2Y1Y2]/2=3.所以-2=3+2b,得b=-5/2.