过抛物线y^2=2px(p>0)的焦点作互相垂直的两条弦,求以这两条弦为对角线的四边形面积的最小值

p>0
F(0.5p,0)
过F互相垂直的两条弦:AB⊥CD
AB或CD⊥X轴,则不符合已知条件,故AB、CD不⊥X轴
设AB:y=k(x-0.5p),x=(y+0.5pk)/k
y^2=2px=2p*(y+0.5pk)/k
ky^2-2py-kp^2=0
(yA+yB)=2p/k,yA*yB=-p^2
(yA-yB)^2=(yA+yB)^2-4yA*yB=(2p/k)^2-4*(-p^2)=4p^2*(1+k^2)/k^2
(xA-xB)^2=(yA-yB)^2/k^2
AB^2=(xA-xB)^2+(yA-yB)^2=(1+1/k^2)*[4p^2*(1+k^2)/k^2]
|AB|=2p(1+k^2)/k^2
k(CD)=-1/k(AB)=-1/k
|CD|=2p(1+k^2)
s=AB,CD为对角线的四边形面积
s=|AB|*|CD|/2
=[2p(1+k^2)/k^2]*[2p(1+k^2)]/2
=2p*[k+(1/k)]^2
k+(1/k)≥2
[k+(1/k)]^2≥4
∴s≥8p
答：以这两条弦为对角线的四边形面积的最小值=8p