Question

已知抛物线y^2=2px上一点P（3/4,√3/2）,且直线x-my-1=0与该抛物线交点为A（x1,y1）,B（x2,

Answer 1

1

y^2=2px

x=3/4 y=√3/2

2p=1

y^2=x

x-my-1=0

y^2=my+1

y^2-my-1=0

y1+y2=m

y1y2=-1

y1=-1/y2

x1=y1^2=1/y2^2=1/x2

y1/x1=(-1/y2)/(1/x2)

x1x2+y1y2=0

向量OA(x1,y1) 向量OB(x2,y2)

OA*OB=0

OA垂直OB

2

S=(1/2)OA*OB=(1/2)√(x1^2+y1^2)*√(x2^2+y2^2)=(1/2)√(x1^2+x1) √(x2^2+x2)

=(1/2)√[(x1x2)^2+x1x2+x1x2^2+x1^2x2] =(1/2)√[2+(x1+x2)]

x1+x2=y1^2+y2^2=y1^2+1/y1^2,(y1-1/y1)^2>=0 y1^2+1/y1^2>=2

S=(1/2)√[2+y1^2+1/y1^2]>=(1/2)√(2+2)=1

y1=1时,S最小=1