Area bounded by the parabola (y - 2)² = x – 1, the tangent to it at the point P (2, 3) and the x-axis is equal to

9 sq. units

$(y-2)^2=(x-1) \Rightarrow 2(y-2) . \frac{d y}{d x}=1$

$\Rightarrow \frac{d y}{d x}=\frac{1}{2(y-2)}$

Thus equation of tangent at P(2, 3) is,

$(y-3)=\frac{1}{2}(x-2)$  i.e. x = 2y - 4

Required area $\Delta=\int_0^3\left((y-2)^2+1-(2 y-4)\right) d y$

$=\left(\frac{(y-2)^3}{3}-y^2+5 y\right)_0^3$ = 9 sq. units