已知过抛物线y^2=2px(p>0)的焦点 斜率为2根号2的直线交抛物线于A(x1,y1),B(x2,y2) -(x10

∵焦点(p/2,0)
∴直线AB：y-0=(2√2)(x-(p/2)),y=(2√2)(x-(p/2))
联立抛物线与直线方程：8(x-(p/2))^2=2px
∴8(x^2-px+(p^2/4))=8x^2-8px+2p^2=2px,8x^2-10px+2p^2=0,4x^2-5px+p^2=0
∴(4x-p)(x-p)=0,x1=p/4,x2=p且x1+x2=5p/4,x1x2=p^2/4
∴|AB|=(√1+k^2)*|x2-x1|=(√1+k^2)*(√(x1+x2)^2-4x1*x2)
=(√9)(√(25p^2/16)-p^2)=3*(√9p^2/16)=3*3p/4=9p/4=9
∴p=4,y^2=8x,直线AB：y=(2√2)*(x-2)；A(1,-2√2),B(4,4√2)
∵向量OA=(1,-2√2),向量OB(4,4√2)
∴向量OC=(1+4λ,λ*(4√2)-(2√2))
∵C在抛物线上
∴[λ*(4√2)-(2√2)]^2=8(1+4λ),32λ^2-2λ*16+8=8+32λ
∴32λ^2-64λ=0,λ=0或2

焦点F(p/2,0)，过焦点的直线方程为y=2√2(x-p/2)，代入抛物线方程y^2=2px并化简得
4x^2-5px+p^2=0 ①
由韦达定理可得
x1+x2=5/4*p ②
x1x2=1/4*p^2 ③
由|AB|=9可得
9=√[1+(2√2)^2]*√[(x1+x2)^2-4x1x2]=3*√[(5/4*p)^2-4*1/4*p^2]=3...