Question

已知抛物线y^2=2px（p>0）的焦点为F，过F作直线l交抛物线于两点A,B求证：|AB|≥2p

Answer 1

焦点F坐标(0.5p,0),设直线L过F,则直线L方程为y=k(x-0.5p)
联立y²=2px得k²x²-(pk²+2p)x+p²k²/4=0
由韦达定理得x1+x2=p+2p/k²
AB=x1+0.5p+x2+0.5p=x1+x2+p=2p+2p/k²=2p(1+1/k²)
因为k=tana,所以1+1/k²=1+1/tan²a
=(sin²a/sin²a)+(cos²a/sin²a)
=(sin²a+cos²a)/sin²a
=1/sin²a
所以AB＝2p（1+1/k²）=2p/sin²a
∵sin²a∈(0,1]
∴AB≥l2pl

Answer 2

焦点F坐标(0.5p,0),设直线L过F,则直线L方程为y=k(x-0.5p)

联立y²=2px得k²x²-(pk²+2p)x+p²k²/4=0

由韦达定理得x1+x2=p+2p/k²

AB=x1+0.5p+x2+0.5p=x1+x2+p=2p+2p/k²=2p(1+1/k²)>2p(k不为0)
当k=0时AB的直线方程为x=0.5p A(0.5p,+-p) |AB|=2p

Answer 3

AB=2P/sin^2(a)（其中a是AB的倾斜角） 满足此条件必须AB过焦点。