JEE Mathematics 2026 Question with Solution

Given: The region is defined by $4x^2 + y^2 \le 8$ and $y^2 \le 4x$.

Find: The area of the common region.

The given curves are:

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

an ellipse, and

$$y^2 = 4x$$

a parabola.

Step 1: Find the points of intersection.

Substitute $y^2 = 4x$ into the ellipse:

$$4x^2 + 4x = 8$$ $$x^2 + x - 2 = 0$$

So,

$$x = 1,\; -2 \quad (\text{only } x=1 \text{ is valid})$$

Thus,

$$y^2 = 4 \Rightarrow y = \pm 2$$

Step 2: Set up the area integral.

For $x \in [0,1]$,

$$\text{Upper curve: } y = \sqrt{4x}, \quad \text{Lower curve: } y = -\sqrt{4x}$$

Area:

$$A = \int_0^1 \left(\sqrt{4x} - (-\sqrt{4x})\right) dx + \text{area under ellipse beyond parabola}$$

Evaluating the integrals gives:

$$A = \pi + \frac{2}{3}$$

Therefore, the area of the region is $\pi + \frac{2}{3}$. The correct option is C.

Given: Two bounding curves, an ellipse $4x^2 + y^2 = 8$ and a parabola $y^2 = 4x$.

Find: The area enclosed by their common region.

The key idea is to identify the intersection first and then decide the effective vertical strip for the common region. From the substitution shown in the working,

$$4x^2 + 4x = 8$$

which gives

$$x = 1,\; -2$$

Only $x = 1$ satisfies the parabola condition for the common region, so the intersection points are $(1,2)$ and $(1,-2)$.

The solution then uses the symmetry about the $x$-axis and the bounded parts of the ellipse and parabola to write the area expression and concludes:

$$A = \pi + \frac{2}{3}$$

Hence, the correct option is C.