(1) solution: let the linear equation be y=k 1x+b and substitute it into the elliptic equation to obtain:

( 1+2k 12)x2+4k 1bx+2 B2-2 = 0，

x 1+x2=-

4k 1b 1+2k 1

2,

The midpoint m is on a straight line,

∴y 1+y22=k 1(

x 1+x22)+b，

So the coordinate of the chord midpoint m is (-

2k 1b 1+2k2，

b 1+2k2)，

k2=-

12k 1，

∴k 1k2=-

12