Let $A=\{(\alpha, \beta) \in \mathbf{R} \times \mathbf{R}:|\alpha-1| \leq 4 \text { and }|\beta-5| \leq 6\}$…

Let
$A=\{(\alpha, \beta) \in \mathbf{R} \times \mathbf{R}:|\alpha-1| \leq 4 \text { and }|\beta-5| \leq 6\}$
and
$B=\left\{(\alpha, \beta) \in \mathbf{R} \times \mathbf{R}: 16(\alpha-2)^2+9(\beta-6)^2 \leq 144\right\}$
  1. $\mathrm{B} \subset \mathrm{A}$
  2. $\mathrm{A} \cup \mathrm{B}=\{(\mathrm{x}, \mathrm{y}):-4 \leq \mathrm{x} \leq 4,-1 \leq \mathrm{y} \leq 11\}$
  3. neither $\mathrm{A} \subset \mathrm{B}$ nor $\mathrm{B} \subset \mathrm{A}$
  4. $A \subset B$

Solution

$\begin{aligned} & \text { A : }|x-1| \leq 4 \text { and }|y-5| \leq 6 \\ & \Rightarrow-4 \leq x-1 \leq 4 \Rightarrow-6 \leq y-5 \leq 6 \\ & \Rightarrow-3 \leq x \leq 5 \quad \Rightarrow-1 \leq y \leq 11 \\ & \text { B : } 16(x-2)^2+9(y-6)^2 \leq 144 \\ & \text { B : } \frac{(x-2)^2}{9}+\frac{(y-6)^2}{16} \leq 1\end{aligned}$

From Diagram $\mathrm{B} \subset \mathrm{A}$

Asked in: JEE Main 2025 (07 Apr Shift 2)

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