Given the boundaries of the area O befound are,

• Curve is (y-1)² = 4 (x+1)

• Line y = x-1

Consider the curve

(y-1)² = 4 (x+1)

Substitute y = x-1

(x-1-1)² = 4(x+1)

(x-2)² = 4x+ 4

x² – 4x + 4 = 4x + 4

x² – 8x = 0

x(x-8) =0

x = 8 (or) x = 0

substituting x in y = x-1

y = 7 (or) y = -1

So , the parabola meets the line y = x-1 at 2 points, D (8,7) and E (0,-1)

As per the given boundaries,

• The parabola (y-1)² = 4 (x+1), with vertex at B(-1,1).

• Line y = x-1

The boundaries of the region to be found are,

•Point B, where the curve (y-1)² = 4 (x+1) has the extreme end the vertex i.e. B (-1,1)

•Point D, where the curve (y-1)² = 4 (x+1) and the line y= x + 1 meet i.e. D (8,7)

•Point E, where the curve (y-1)² = 4 (x+1) and the line y = x-1 meet i.e. E (0,-1)

•Point O, the origin

Consider the parabola,

(y-1)² = 4(x+1)

Area of the required region = Area of EABCDE

Area of EABCDE = Area above line ED - Area above EABCD

[Using the formula ]

The Area of the required region