△BOC~△PAC, OB:AP=OC:AC, so AP=4 ∴P(- 1,4),∴k=-4.

(2)

The point below the X axis is easy to find. The intersection of the straight line BC and the hyperbola is (2, -2), so 0.

Above the X axis (I'm going to eat first, and we'll discuss it later if I don't understand)

Let's put the coordinates of c', the symmetry point of c ABout ab, and then ∠C'BA=ABC.

The intersection of BC' line and hyperbola is the lowest end of M (unavailable), and M should be above this point (note that a < 0).