Question

已知过抛物线y²=2px（p>0)的焦点F的直线与抛物线相交于M,N两点自准线l作垂线,垂足分别为M1,N1

Answer 1

结论成立.

证明：设过F点的直线：x=ty+p/2联立y^2=2px 得y^2-2pty-p^2=0

y1+y2=2pt,

y1y2=-p^2

x1+x2=ty1+ty2+p=t(y1+y2)+p=t(2pt)+p=2pt^2+p

x1x2=(ty1+p/2)(ty2+p/2)=t^2y1y2+pt(y1+y2)/2+p^2/4=p^2/4

(S2)^2=(1/2|M1N1|*P)^2=1/4[|Y1-Y2|^2*P^2]=1/4[(Y1+Y2)^2-4Y1Y2]*P^2

=1/4[4(pt)^2+4p^2]*p^2=(t^2+1)p^4 (1)

4S1S3=4*(1/2*Y1*(P/2+X1))*(1/2*(-Y2)*(P/2+X2))=-Y1Y2*(P^2/4+(X1+X2)P/2+X1X2)

=p^2[p^2/4+(2pt^2+p)p/2+p^2/4]=(t^2+1)p^4 (2)

(1)=(2) 所以结论成立