LetA=\ (α, β) ∈ R × R :|α-1| ≤ 4 and |β-5| ≤ 6\ and B=\ (α, β) ∈ R × R : 16(α-2)^2+9(β-6)^2 ≤ 144\

Correct answer: A. B A aligned & A : |x-1| ≤ 4 and |y-5| ≤ 6 & -4 ≤ x-1 ≤ 4 -6 ≤ y-5 ≤ 6 & -3 ≤ x ≤ 5 -1 ≤ y ≤ 11 & B : 16(x-2)^2+9(y-6)^2 ≤ 144 & B : ((x-2)^2)/(9)+((y-6)^2)/(16) ≤ 1 aligned From Diagram B A

JEE Main 20257 AprilEvening Shift

Exam: JEE Main
Year: 2025
Date: 7 April
Shift: Evening Shift
Subject: Mathematics
Chapter: Ellipse

JEE Main 2025 · 7 April · Evening Shift · Mathematics · Ellipse

LetA=\ (α, β) ∈ R × R :|α-1| ≤ 4 and |β-5| ≤ 6\ and B=\ (α, β) ∈ R...

LetA=\{(\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\}

Options

A $\mathrm{B} \subset \mathrm{A}$
B $\mathrm{A} \cup \mathrm{B}=\{(\mathrm{x}, \mathrm{y}):-4 \leq \mathrm{x} \leq 4,-1 \leq \mathrm{y} \leq 11\}$
Cneither $\mathrm{A} \subset \mathrm{B}$ nor $\mathrm{B} \subset \mathrm{A}$
D $A \subset B$

Correct answer

A. $\mathrm{B} \subset \mathrm{A}$

Step-by-step 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}

Practice Ellipse on Quantrex Academy →

More from Ellipse