JEE Mathematics 2024 Question with Solution

Given: The parabola is $(y-2)^2 = x-1$ and the line is $x-2y+4 = 0$. The region lies in the first quadrant and is enclosed with the positive coordinate axes.

Find: The area of the enclosed region.

Write both curves as $x$ in terms of $y$:

$$x = (y-2)^2 + 1$$ $$x = 2y - 4$$

Now find the key intersection points shown in the solution.

Intersection of parabola and line:

$$(y-2)^2 + 1 = 2y - 4$$ $$y^2 - 4y + 4 + 1 = 2y - 4$$ $$y^2 - 6y + 9 = 0$$ $$(y-3)^2 = 0$$

So,

$$y = 3,\, x = 2$$

Hence the curves touch at $(2,3)$.

Intersection of the line with the positive $y$-axis:

$$x = 0 \implies 2y - 4 = 0 \implies y = 2$$

So the point is $(0,2)$.

Intersection of the parabola with the positive $x$-axis:

$$y = 0 \implies x = (0-2)^2 + 1 = 5$$

So the point is $(5,0)$.

The area is split into two parts.

For $0 \le y \le 2$, the region is between the parabola and the $y$-axis:

$$A_1 = \int_0^2 \left(((y-2)^2 + 1) - 0\right) \, dy$$ $$A_1 = \int_0^2 (y^2 - 4y + 5) \, dy$$ $$A_1 = \left[ \frac{y^3}{3} - 2y^2 + 5y \right]_0^2$$ $$A_1 = \frac{8}{3} - 8 + 10 = \frac{14}{3}$$

For $2 \le y \le 3$, the region is between the parabola and the line:

$$A_2 = \int_2^3 \left(((y-2)^2 + 1) - (2y - 4)\right) \, dy$$ $$A_2 = \int_2^3 (y^2 - 6y + 9) \, dy$$ $$A_2 = \int_2^3 (y-3)^2 \, dy$$ $$A_2 = \left[ \frac{(y-3)^3}{3} \right]_2^3 = \frac{1}{3}$$

Therefore,

$$A = A_1 + A_2 = \frac{14}{3} + \frac{1}{3} = 5$$

Therefore, the area of the required region is $5$ square units.

Given: The parabola is $(y-2)^2 = x-1$ and the line is $x-2y+4 = 0$.

Find: The enclosed area in the first quadrant.

From the extracted working,

$$(y - 2)^2 = x - 1 \implies y - 2 = \sqrt{x - 1}$$

and

$$x - 2y + 4 = 0 \implies y = \frac{x + 4}{2}$$

Equating them,

$$\frac{x+4}{2} = \sqrt{x-1} + 2$$

Rearranging and squaring gives

$$(x-2)^2 = 0$$

Hence,

$$x = 2$$

and then

$$y = \frac{2+4}{2} = 3$$

So the intersection point is $(2,3)$.

The extracted solution concludes the area evaluates to $5$. The more geometrically consistent setup is to integrate with respect to $y$ as shown in the standard method above, which gives the same final value.

Therefore, the required area is $5$.