焦點(diǎn)F(p/2,0),過(guò)焦點(diǎn)的直線方程為y=2√2(x-p/2),代入拋物線方程y^2=2px并化簡(jiǎn)得
4x^2-5px+p^2=0①
由韋達(dá)定理可得
x1+x2=5/4*p ②
x1x2=1/4*p^2③
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*3/4*p=9/4*p
得p=4.于是拋物線方程為y^2=8x
向量OC=向量OA＋λOB,若C點(diǎn)與A點(diǎn)重合,則λ=0；
若C點(diǎn)相異于A點(diǎn),則C點(diǎn)必為過(guò)A點(diǎn)且平行于OB的直線與拋物線的交點(diǎn).
將p=4代入方程①,有4x^2-20x+16=0,解得x1=1,x2=4.依題意有y1=-2√2,y2=4√2.于是有
A(1,-2√2),B(4,4√2).則直線OB的斜率為4√2/4=√2,于是直線AC的方程為y+2√2=√2(x-1),即
y=√2(x-3),代入y^2=8x,可得(x-1)(x-9)=0,解得x=1或x=9.顯然x=1對(duì)應(yīng)點(diǎn)A(1,-2√2),x=9對(duì)應(yīng)點(diǎn)C(9,6√2).于是λ=AC/OB=(6√2-1)/(4-0)=(6√2-1)/4（因?yàn)樾甭识际恰?,故線段長(zhǎng)度之比等于橫坐標(biāo)差之比）