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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
人气:305 ℃ 时间:2020-05-25 22:05:49
解答
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...
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