设A(x1,y1),B(x2,y2) 且x1 x1+x2=4-2√2=A
②－①:x2-x1=-(y2-y1)
ax1²+by1²=1 ③
ax2²+by2²=1 ④
④－③:a(x2²-x11²)+b(y2²-y1²)=0 => a(x1+x2)(x2-x1)+b( y1+y2)(y2-y1)=0,带入已知量,得：
(x2-x1)A(a-b/√2)=0,所以b=√2a
|AB|=2√2,(x2-x1)²+(y2-y1)²=2(x2-x1)²=(2√2)²,所以x2-x1=2,又x1+x2=4-2√2,则x1=1-√2,x2=3-√2,
y1=1-x1=√2,带入椭圆方程：ax1²+√2ay1²=1 =>
a=1/3,b=√2a=√2/3