The area of the region bounded by the parabola (y – 2)2 =x – 1, the tangent to it at the point with the ordinate 3 and the x-axis is
The question describes something like –
To solve the problem, we need to find the points 1, 2, and 3 first. They are the key to setting up the bounds of our integration and understanding what function to integrate.
For point 1:-
We know it is a point on the parabola with ordinate 3, at which the tangent to the parabola is taken.
Plugging in y = 3 into the equation for the parabola (y – 2)2 = x – 1,
(3 – 2)2 = x – 1
⇒ 1 = x – 1
⇒ x = 2
So, Point 1 is (2, 3)
For point 2:-
We need to find the point of intersection of the parabola and the x – axis.
We know that ordinate at this point is 0.
Plugging in y = 0 into the equation for the parabola (y – 2)2 = x – 1,
(0 – 2)2 = x – 1
⇒ 4 = x – 1
⇒ x = 5
So, Point 2 is (5, 0)
For point 3:-
Tangent at any point (x1, y1) for a curve is –
y – y1 = (x – x1)
For parabola (y – 2)2 = x – 1, differentiating both sides of the equation –
2(y – 2) = 1
⇒
So, slope at point 1, i.e., (2, 3) is –
So, equation of tangent at (2, 3) is –
y – 3 = � (x – 2)
⇒ 2y – 6 = x – 2
⇒ x – 2y + 4 = 0
The tangent intersects the x – axis at point 3. At this point, ordinate is 0.
So, plugging y = 0 in the equation of the tangent x – 2y + 4 = 0, -
x – 2 �0 + 4 = 0
⇒ x = -4
So, point 3 is (-4, 0).
Now, that we have the 3 points, let’s figure out how to compute the area required.
The area we need can be divided into 3 sections –
i. x = -4 to x = 1
Here, area required = area enclosed by the tangent and the x – axis
ii. x = 1 to x = 2
Here, area required = (area enclosed by the tangent and the x – axis) – (area enclosed within the parabola)
iii. x = 2 to x = 5
Here, area required = area enclosed by the parabola and the x – axis
The parabola is (y – 2)2 = x – 1
Solving for y,
y = 2 ± √(x – 1)
So, area A we need is –
(Ans)