过P(0,2)的直线交椭圆x2/2+y2=1于A,B若三角形AOB面积为2/3求AB方程

Question

过P(0,2)的直线交椭圆x2/2+y2=1于A,B若三角形AOB面积为2/3求AB方程

Answer 1

设直线为y-2=kx
与椭圆方程X^2/2+Y^2=1联立,得到
X^2/2+(kx+2)^2=1
(k^2+1/2)x^2+4kx+3=0
根的判别式=16k^2-12(k^2+1/2)=4k^2-6>0
k^2>2/3
x1+x2=-4k/(k^2+1/2)
x1x2=3/(k^2+1/2)
两交点距离=根号下(k^2+1)乘以|x2-x1|
=根号下(k^2+1)乘以|x2-x1|
原点到两交点所在直线距离
=|k*0-0+2|/根号下(k^2+1)
三角形面积
=1/2*根号下(k^2+1)乘以|x2-x1|乘以|k*0-0+2|/根号下(k^2+1)
=|x2-x1|
=根号下[(x1+x2)^2-4x1x2]
=根号下[(4k^2-6)/(k^2+1/2)^2]
=根号下[4/(k^2+1/2)-8/(k^2+1/2)^2]
=2/3
换元,令m=4/(k^2+1/2),m2/3.的条件.