已知抛物线y²=2px(p＞0),过点A(0,-1)作直线l与抛物线相交于P,Q两点,点B的坐标为(0,1),

设l:x=m(y+1),①
代入y^2=2px,得y^2-2mpy-2mp=0,
设P(x1,y1),Q(x2,y2),则y1+y2=2mp,y1y2=-2mp,
由①,x1x2=m(y1+1)*m(y2+1)=m^2*(y1y2+y1+y2+1)=m^2,
△=4m^2p^2+8mp,|y1-y2|=√△,
BP的斜率k1=(y1-1)/x1,BQ的斜率k2=(y2-1)/x2,
依题意k1k2=(y1-1)(y2-1)/(x1x2)=-3,
∴(y1-1)(y2-1)=-3x1x2,
∴0=3x1x2+y1y2-(y1+y2)+1
=y1y2-(y1+y2)+3m^2+1
=-4mp+3m^2+1,
∴mp=(3m^2+1)/4,△=(3m^2+1)(3m^2+9)/4,
画示意图知,tan∠MBN=|(k2-k2)/(1+k2k1)|=|[(y2-1)/x2-(y1-1)/x1]/[1-3]|
=(1/2)|m[(y1+1)(y2-1)-(y2+1)(y1-1)]/(x1x2)|
=|y1-y2|/|m|=√[(3m^2+1)(3m^2+9)]/(2|m|),
无法求出角的大小,请检查题目