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 =(