Question

已知抛物线yﾁ0ﾅ5=2px,过点c(-2,0)的直线l交抛物线于a,b两点,oa×ob=12

Answer 1

AB:x=my-2
y^2=2px=2p*(my-2)
y^2-2pmy+4p=0
yA+yB=2pm,yA*yB=4p
xA*xB=(myA-2)*(myB-2)=4
xA*yB+xB*yA=(myA-2)*yB+(myB-2)*yA=2m(yA*yB)-2(yA+yB)=4pm
(xA^2+yA^2)*(xB^2+yB^2)
=(xA*xB)^2+(yA*yB)^2+(xA*yB+xB*yA)^2-2xA*xB*yA*yB
=4^2+(4p)^2+(4pm)^2-2*4*4p
=16-16p+16p^2+16p^2*m^2
OA*OB=12
OA^2*OB^2=144
16-16p+16p^2+16p^2*m^2=144
m^2=(8+p-p^2)/p^2
题目所什么没有写好，只能到这一步。