“*”是“×”
设A(xA,yA),B(xB,yB)且xB≥xA
联立C与l,得
3x^2+4(x/2+m)^2-12=0
x^2+mx+m^2-3=0
其解为xA与xB,
yA=xA/+m,yB=xB/2+m
l(PA):y-yA=k(x-xA)
l(PB):y-yB=-k(x-xB)
联立l(PA)与l(PB),得
k*(x-xA)+yA=-k*(x-xB)+yB
x=(k*xA+k*xB+yB-yA)/(2k)
=[k*(xA+xB)+(xB/2+m)-(xA/2+m)]/(2k)
=[-k*m+(xB-xA)/2]/(2k)
={-k*m+[(xA+xB)^2-4*xA*xB]^(1/2)/2}/(2k)
={-k*m+[m^2-4*(m^2-3)]^(1/2)/2}/(2k)
=[-2k*m+(-3*m^2+12)^(1/2)]/(4k)
恒有x与m有关,故我认为,是我认为,不存在这样的P
现在该我请教你了,会做了之后,有劳你什么时候有空告诉我