解法一:(1)若θ=
,此时|AB|=2p.
(2)若θ≠
,
∵有两个交点,∴θ≠0.
设AB:y=k(x-
),代入抛物线方程化为k2x2-(k2p+2p)x+
=0.
∴x1+x2=
,x1x2=
,
|AB|=
=2p·
=2p(1+
)＞2p.
∴|AB|min=2p.此时,θ=
.
解法二:设直线AB的方程为x=my+
,代入抛物线方程有y2-2pmy-p2=0.
∴y1+y2=2pm,y1y2=-p2.
∴|y1-y2|=
=
.
∴|AB|=
|y1-y2|=
·2p
=2p(1+m2).
∵m∈R,
∴|AB|=2p(1+m2)≥2p.
此时m=0,即θ=
时,|AB|min=2p.
解法三:AB是焦点弦.根据抛物线的定义,|AB|等于点A与点B分别到其准线的距离之和.而准线为x=-
,∴|AB|=d1+d2=(x1+
)+(x2+
)=x1+x2+p(以下同解法一).