∴|xp|×|yp|=|k|= 1，

∴ Let the coordinate of p be (a, 1/a )(a is a positive number),

∵PA⊥x axis,

∴ the abscissa of a is a,

∫A is on y=-2/x,

∴A's coordinate is (a, -2/a).

∵PB⊥y axis,

The ordinate of ∴B is 1 /a,

∫B is on y=-2 /x,

∴ substitution: 1/a=-2/x

Solution: x=-2a,

The coordinate of ∴B is (-2a, 1 /a),

∴PA=| 1/a -(-2 /a )|=3 /a

，PB=|a-(-2a)|=3a，

∵PA⊥x axis, PB⊥y axis, x axis, ⊥y axis,

∴PA⊥PB，

The area of ∴△PAB is:1/2pa× Pb =1/2× 3/a× 3a = 9/2.