設(shè)A點坐標(biāo)(x1,y1),B點坐標(biāo)(x2,y2)
將x+y=1代入mx²+ny²=1,得(m+n)x²-2nx+n-1=0,(易知m,n＞0)
根據(jù)韋達定理有
x1+x2=2n/(m+n),x1x2=(n-1)/(m+n)
故y1+y2=2-(x1+x2)=2m/(m+n)
AB=√[(x1-x2)²+(y1-y2)²]=√[(x1-x2)²+(1-x1-1+x2)²]
=√[2(x1-x2)²]=√{2[(x1+x2)²-4x1x2]}
=√{2[4n²/(m+n)²-4(n-1)/(m+n)]}
=√[8(m-mn+n)/(m+n)²]=2√2
變形得m-mn+n=(m+n)²
C點坐標(biāo)為((x1+x2)/2,(y1+y2)/2)
故[(y1+y2)/2]/[(x1+x2)/2]=(y1+y2)/(x1+x2)=m/n=√2/2
即m=(√2/2)n,代入m-mn+n=(m+n)²得
(√2/2)n-(√2/2)n²+n=[(√2/2)n+n]²
(√2+2)(√2-1)=3n
n=√2/3
m=1/3
故橢圓方程為(1/3)x²+(√2/3)y²=1