The given parabola is (y−2)2=x−1 Vertex (1,2) and it meets x-axis at (5,0) Also it gives y2−4y−x+5=0 So, that equation of tangent to the parabola at (2,3) is y.3−2(y+3)−‌
1
2
(x+2)+5=0 or x−2y+4=0 which meets x-axis at (−4,0).
In the figure shaded area is the required area.
Let us draw PD perpendicular to y-axis.
Then required area ‌=Ar∆BOA+Ar(OCPD)−Ar(△APD) ‌=‌