Given curve : y2−2x−2y=1. Can be written as (y−1)2=2(x+1)
And, the given information can be plotted as shown in figure Tangent at A:2y−x−5=0 using T=0} Intersection with y=1 is x=−3 Hence, point P is (−3,1) Taking advantage of symmetry Area of △PAB=2×