Area Under the Curve
Area — Ellipse Intersected with Parabola Region
nta_pyq_2026_jan
Grade 12
Question:
The area of the region $A=\{(x,y):4x^2+y^2\leq8\text{ and }y^2\leq4x\}$ is:
$\pi+4$
$\pi+\dfrac{2}{3}$
$\dfrac{\pi}{2}+2$
$\dfrac{\pi}{2}+\dfrac{1}{3}$
Step-by-Step Solution
Key Concept: Ellipse $\frac{x^2}{2}+\frac{y^2}{8}=1$ and parabola $y^2=4x$. Intersection: $4x^2+4x=8\Rightarrow x=1$ (first quadrant). Split the area: parabolic region from $x=0$ to $1$, then ellipse region from $x=1$ to $\sqrt{2}$, doubled for symmetry.
Area $=\pi+2/3$.
Correct Answer: 2