Simultaneous equation:

{mx^2+ny^2= 1

{y=-x+ 1

Eliminate y and get:

mx^2+n(-x+ 1)^2= 1

Finishing, you must:

(m+n)x^2-2nx+n- 1=0

Let A(x 1, y 1) and B(x2, y2).

From Vieta's theorem, I get:

x 1+x2=2n/(m+n)

x 1x2=(n- 1)/(m+n)

∵ The abscissa of point m in AB is 2-√ 2.

∴(x 1+x2)/2=2-√2

X 1+x2 = 4-2 √ 2 = 2n/(m+n) ①.

| ab | 2 = ( 1+k 2)？ 【(x 1+x2)^2-4x 1x2]=2？ [24- 16√2-4x 1x2]=8

X1x 2 = 5-4 √ 2 = (n-1)/(m+n) ②.

Simultaneous ① ②, the solution is:

{m= 1/3

{n=(√2)/3

The elliptic equation is x 2/3+(√ 2/3) y 2 =1.