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 figureTangent at A:2y−x−5=0 using T=0}Intersection with y=1 is x=−3Hence, point P is (−3,1)Taking advantage of symmetryArea of △PAB=2×21×(1−(−3))×(3−1)=8 sq. units