JEE Mathematics 2024 Question with Solution

Given: The parabola is $y^2 = 4(x - 2)$ and the line is $y = 2x - 8$.

Find: The area of the region bounded by these two curves.

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

$$x = \frac{y^2}{4} + 2$$

and from $y = 2x - 8$,

$$x = \frac{y + 8}{2} = \frac{y}{2} + 4$$

Now find the points of intersection by substituting $x = \frac{y+8}{2}$ into the parabola:

$$y^2 = 4\left(\frac{y + 8}{2} - 2\right)$$ $$y^2 = 2(y+4)$$ $$y^2 - 2y - 8 = 0$$ $$(y-4)(y+2)=0$$

So, the curves intersect at $y = 4$ and $y = -2$.

For $y \in [-2,4]$, the line lies to the right of the parabola, so the required area is

$$A = \int_{-2}^{4} \left[\left(\frac{y}{2}+4\right) - \left(\frac{y^2}{4}+2\right)\right] \, dy$$ $$A = \int_{-2}^{4} \left(-\frac{y^2}{4} + \frac{y}{2} + 2\right) \, dy$$

Integrating,

$$A = \left[-\frac{y^3}{12} + \frac{y^2}{4} + 2y\right]_{-2}^{4}$$

At $y=4$,

$$-\frac{4^3}{12} + \frac{4^2}{4} + 2(4) = -\frac{16}{3} + 12 = \frac{20}{3}$$

At $y=-2$,

$$-\frac{(-2)^3}{12} + \frac{(-2)^2}{4} + 2(-2) = \frac{2}{3} + 1 - 4 = -\frac{7}{3}$$

Therefore,

$$A = \frac{20}{3} - \left(-\frac{7}{3}\right) = \frac{27}{3} = 9$$

Hence, the area of the bounded region is $9$ square units. The correct option is B.

Given: $y^2 = 4(x - 2)$ and $y = 2x - 8$.

Find: Area enclosed between the parabola and the line.

Using the line equation,

$$y = 2x - 8$$ $$2x = y + 8$$ $$x = \frac{y+8}{2}$$

Substitute this in the parabola equation:

$$y^2 = 4(x-2)$$ $$y^2 = 4\left(\frac{y+8}{2} - 2\right)$$ $$y^2 = 2y + 8$$ $$y^2 - 2y - 8 = 0$$ $$(y-4)(y+2)=0$$

Thus, the intersection limits are $y=-2$ and $y=4$.

Now,

$$x_{\text{line}} = \frac{y}{2} + 4$$ $$x_{\text{parabola}} = \frac{y^2}{4} + 2$$

So,

$$A = \int_{-2}^{4} (x_{\text{line}} - x_{\text{parabola}}) \, dy$$ $$A = \int_{-2}^{4} \left(\frac{y}{2}+4-\frac{y^2}{4}-2\right) \, dy$$ $$A = \int_{-2}^{4} \left(-\frac{y^2}{4} + \frac{y}{2} + 2\right) \, dy$$

Now evaluate termwise:

$$\int_{-2}^{4} -\frac{y^2}{4} \, dy = -6$$ $$\int_{-2}^{4} \frac{y}{2} \, dy = 3$$ $$\int_{-2}^{4} 2 \, dy = 12$$

Hence,

$$A = -6 + 3 + 12 = 9$$

Therefore, the enclosed area is $9$ square units, so the correct option is B.