I think your problem instance differs from your graph: Let us, for instance, choose p=0.0 . Then the constraint reads x_2 <= 0. Thus, your feasible set in the graph of approx P={(x,y) in R^2 : y <= 1-x^2} contains the point (x,y)=(0,1), whereas the equations permit this point.
Hey again :) I corrected your statement: It must be 2*p * x_1 + x_2 <= 1 + p^2 . — Preceding unsigned comment added by 92.213.9.228 (talk) 16:24, 15 May 2016 (UTC)
