Given, the parabola (y−2)2=(x−1) and y=3, x=2
So, point is (2,3) Differentiate given equation w.r.t. x, we get
2(y−2)y′=1
y′=2(y−2)1
y′∣(2,3)=21
So, the equation of tangent at (2,3)
y−3=21(x−2)
x−2y+4=0

Required area =∫03[(y−2)2+1]dy−∫03(2y−4)dy
Required area =∫03(y2−4y+5)dy+∫03(2y−4)dy
Required area =[3y3−2y2+5y]03−[y2−4y]03
Required area=9sq units