若抛物线y2=2px(p>0)与直线x-y-1=0相交于A,B两点,且OA•OB=-1,则p=( ) A.1 B.2 C.4 D.8
问题描述:
若抛物线y2=2px(p>0)与直线x-y-1=0相交于A,B两点,且
•OA
=-1,则p=( )OB
A. 1
B. 2
C. 4
D. 8
答
设A(x1,y1),B(x2,y2),则y2=2pxx−y−1=0消去y,得:(x-1)2=2px,即x2-(2+2p)x+1=0,∴x1+x2=2+2p,x1x2=1,∴y1y2=(x1-1)(x2-1)=x1x2-(x1+x2)+1=1-(2+2p)+1=-2p,∵OA•OB=-1,∴x1x2+y1y2=-1...