Let $\mathrm{A}=\{(\alpha, \beta) \in \mathbf{R} \times \mathbf{R}:|\alpha-1| \leq 4$ and $|\beta-5| \leq 6\}$ and $B=\left\{(\alpha, \beta) \in \mathbf{R} \times \mathbf{R}: 16(\alpha-2)^{2}+9(\beta-6)^{2} \leq 144\right\}$. Then

(A) $\mathrm{B} \subset \mathrm{A}$

(B) $\mathrm{A} \cup \mathrm{B}=\{(\mathrm{x}, \mathrm{y}):-4 \leq \mathrm{x} \leq 4,-1 \leq \mathrm{y} \leq 11\}$

(C) neither $\mathrm{A} \subset \mathrm{B}$ nor $\mathrm{B} \subset \mathrm{A}$

(D) $\mathrm{A} \subset \mathrm{B}$

JEE Main · 2025 · 07 Apr 2025 · Shift 2 · Ellipse

Let

A = {(α, β) \in R \times R : ∣ α - 1∣ \leq 4

and

∣ β - 5∣ \leq 6}

and

B = {(α, β) \in R \times R : 16 (α - 2)^{2} + 9 (β - 6)^{2} \leq 144}

.
Then

Answer: (A)

The conditions

∣ x - 1∣ \leq 4

and

∣ y - 5∣ \leq 6

describe set

A

as the rectangle

- 3 \leq x \leq 5, - 1 \leq y \leq 11

. The condition

16 (x - 2)^{2} + 9 (y - 6)^{2} \leq 144

rewrites as

\frac{( x - 2 ) ^{2}}{9} + \frac{( y - 6 ) ^{2}}{16} \leq 1

, an ellipse centred at

(2, 6)

with horizontal semi-axis

3

and vertical semi-axis

4

. Hence set

B

lies within

- 1 \leq x \leq 5

and

2 \leq y \leq 10

. Both ranges sit inside the rectangle, and since the ellipse is contained in its bounding box, every point of

B

belongs to

A

. Thus

B \subset A

JEE Main · 2025 · 07 Apr 2025 · Shift 2 · Single correct option