如图直线aby=kx+2k+4与抛物线y=1/2x2交于AB两点.当k＝一1／2时,在直线AB下的抛物线上求一点P,使三角形ABP的面积为5

Question

如图直线aby=kx+2k+4与抛物线y=1/2x2交于AB两点.当k＝一1／2时,在直线AB下
的抛物线上求一点P,使三角形ABP的面积为5

Answer 1

SΔABP=SΔPBA+SΔPQB
=1/2PQ*AC+1/2PQ*BD
=1/2PQ×5=10,
∴PQ=2,
设P(m,1/2m^2),则Q(m,-1/2m+3),

∴PQ=-1/2m+3-1/2m^2=2,
m^2+m-2=0,
(m+2)(m-1)=0,
m=-2或1,
∴P(-2,2)或(1,1/2).