Area Under The Curves Part 1

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)
‌=‌

1

2

×4×2+

3

∫

0

xdy−‌

1

2

×2×1
‌=3+

3

∫

0

(y−2)2+1dy
‌=3+[‌

(y−2)3

3

+y]03
‌=3+[‌

1

3

+3+‌

8

3

]=3+6=9‌ Sq. units ‌