1.C(- 1, 6) is on y=k/x, so 6=k/(- 1), so k=-6, so the hyperbola is y=-6/x,

Because A(-2, n) is also on y=-6/x, so n=-6/(-2)=3, so A(-2, 3),

Let a straight line be y=kx-3 and substitute it into (-2,3) to get k=-3, then the analytical formula of AB is y=-3x-3.

2. Let the straight line DE be y=kx-3/2, and substitute (1, 0), so k=3/2, so DE is y=3/2x-3/2, which is obtained by combining with straight line AB.

F(- 1/3, -2), because EB=3-3/2=3/2, the area of △BDF is1/2 * (3/2) *1+1/2 * (.

Let the straight line DE be y=kx-m and substitute it into (1, 0), so k=m, so DE is y=mx-m, which is obtained by combining with the straight line AB.

F(m-3/m+3, -6m/m+3), because m

So EB=-m+3, so the area of △BDF is1/2 * (3-m) *1+1/2 * (3-m) * [-(m-3/m+3)] = (9.