Parabola Part 1

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×‌

1

2

×(1−(−3))×(3−1)
=8 sq. units