题目

椭圆x^2/9 + y^2/3=1与直线y=kx-2交A,B两点,P(0,1),且|PA|=|PB|,求直线方程

提问时间：2020-12-27

答案

x^2/9 + y^2/3=1,即x^2+3y^2-9=0
将y=kx-2代入得：x^2+3(kx-2)^2-9=0
(3k^2+1)x^2-12kx+3=0
x1+x2=12k/(3k^2+1)
P(0,1）
令A(x1,y1),B(x2,y2)
PA^2=x1^2+(y1-1)^2,PB^2=x2^2+(y2-1)^2
|PA|=|PB|,PA^2=PB^2
x1^2+(y1-1)^2=x2^2+(y2-1)^2
(x1+x2)(x1-x2)+(y1+y2-2)(y1-y2)=0
y=kx-2
(x1+x2)(x1-x2)+(kx1+kx2-6)(kx1-kx2)=0
(x1-x2){(x1+x2)+k^2(x1+x2)-6k} = 0
A,B不在y轴上,所以x1-x2≠0
∴(x1+x2)+k^2(x1+x2)-6k = 0
(k^2+1)(x1+x2)-6k=0
(k^2+1)*12k/(3k^2+1)-6k=0
12k^3+12k-18k^3-6k=0
6k^3-6k=0
k(k+1)(k-1)=0
k=0时,与椭圆无交点
∴k=-1,或1
直线方程y=-x-2,或y=x-2