Area Under The Curves Part 1

Given parabola
(y−2)2=(x−1)
Since, Ordinate = y = 3
Then, x = 2
Point on parabola (2, 3)
Differentiating Eq. (i) w.r.t. x, we get
2(y−2)

dy

dx

=1

dy

dx

=

1

2(y−2)

At (2, 3)

dy

dx

=

1

2

Equation of tangent at (2, 3)
y−3=

1

2

(x−2)
or x − 2y + 4 = 0
Intersection point of parabola on X-axis is
y = 0, x = 5 i.e (5, 0)
Intersection point of tangent and X-axis
y = 0, x = − 4 i.e (− 4, 0)
Area of shaded region =∫03[(y−2)2+1−(2y−4)]dy
=∫03(y2−6y+9)dy
=(

y3

3

−3y2+9y)03=9 sq. unit.