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) -(x1求该抛物线的方程O为坐标原点,C为抛物线上一点,若向量OC=向量OA+λOB,求λ的值
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) -(x1
O为坐标原点,C为抛物线上一点,若向量OC=向量OA+λOB,求λ的值
|AB|=x1+p/2+x2+p/2=x1+x2+p (x1+x2)=9-p
|AB|=√(k^2+1)|x1-x2| =3|x1-x2| =9
(x1-x2)^2=9
y=k(x-p/2)
k^2(x^2-px+p^2/4)=2px
k^2x^2-(k^2p+2p)x+k^2p^2/4=0
x1x2=p^2/4
(x1+x2)^2=(9-p)^2=(x1-x2)^2+4x1x2=9+p^2
81-18p+p^2=9+p^2
72-18p=0
p=4
方程为y^2=8x
[[[1]]]
y²=8x
[[[2]]]
λ=0或λ=5
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*3/4*p=9/4*p
得p=4.于是抛物线方程为y^2=8x
向量OC=向量OA+λOB,若C点与A点重合,则λ=0;
若C点相异于A点,则C点必为过A点且平行于OB的直线与抛物线的交点.
将p=4代入方程①,有4x^2-20x+16=0,解得x1=1,x2=4.依题意有y1=-2√2,y2=4√2.于是有
A(1,-2√2),B(4,4√2).则直线OB的斜率为4√2/4=√2,于是直线AC的方程为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.显然x=1对应点A(1,-2√2),x=9对应点C(9,6√2).于是λ=AC/OB=(6√2-1)/(4-0)=(6√2-1)/4 (因为斜率都是√2,故线段长度之比等于横坐标差之比)
(1)y²=8x
(2)λ=0,或λ=2