Question

已知中心在原点，焦点在坐标轴上的椭圆与直线x+y=1相交于A,B两点，

且AB绝对值=2√2，，AB的中点C与椭圆中心O的连线的斜率为√2/2，求椭圆的方程。

Answer 1

设A(x1,1-x1)，B(x2,1-x2)，C((x1+x2)/2，(1-(x1+x2)/2)
AB^2=(x2-x1)^2+(x2-x1)^2=8 ，x2-x1=2或x2-x1=-2
(1-(x1+x2)/2)/((x1+x2)/2)=√2/2 x1+x2=4-2√2
x2-x1=2
x1+x2=4-2√2
x1=1-√2，y1=1-x1=√2
x2=3-√2，y2=1-x2=1-3+√2=√2-2
x2-x1=-2
x1+x2=4-2√2
x1=3-√2，y1=1-x1=1-3+√2=√2-2
x2=1-√2，y2=1-x2=1-1+√2=√2

椭圆：
(1-√2)^2/a^2+(√2)^2/b^2=1
(3-√2)^2/a^2+(√2-2)^2/b^2=1
可求出a、b

Answer 2

mx^2+ny^2=1
x+y=1
mx^2+n(1-x)^2=1
(m+n)x^2-2nx+n^2-1=0
x1+x2=2n/(m+n)
x1x2=(n^2-1)/(m+n)
(x1-x2)^2=(x1+x2)^2-4x1x2
=4n^2/(m+n)^2-4(n^2-1)/(m+n)
|AB|^2=2(x1-x2)^2=8
Cx=(x1+x2)/2=n/(m+n) Cy=1-n/(m+n) =m/(m+n)
Cy/Cx=m/n=√2/2
m=√2/2n
(x1-x2)^2=4 n^2/(m+n)^2-(n^2-1)/(m+n)=1
1/(1+√2/2)-(n^2-1)/(1+√2/2)n=1
n-(n^2-1)=(1+√2/2)n
1-n^2=√2n/2
(n+√2/4)^2=1+1/8
n+√2/4=3/√8
n=√2/2
m=1/2
x^2/2+y^2/√2=1

Answer 3

(x²/3)+[(√2)y²/3]=1