Let A = {(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}. Then:

 ❓ Question

Let

$$A=\{(\alpha , \beta ) \in \; R\; \times \; R:\mathrm{\mid }\mathrm{\alpha }- 1\mathrm{\mid }\le 4,\mathrm{\mid }\mathrm{\beta }- 5\mathrm{\mid }\le 6\}$$

and

$$B=\{(\alpha , \beta ) \in R\; \times \; R:16(\mathrm{\alpha }- 2{)² }^{}+ 9(\mathrm{\beta }- 6{)² }^{}\le 144\}\mathrm{.}$$

Then determine the relationship between sets $A$ and $B$ (containment, intersection, union, etc.).

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🖼️ Question Image

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✍️ Short Solution

Step 1 — Convert descriptions into ranges/standard form

• For $A$:
  $|\alpha-1|\le4 \Rightarrow \alpha\in[-3,5]$
  $|\beta-5|\le6 \Rightarrow \beta\in[-1,11]$
  So $A$ is an axis-aligned rectangle with opposite corners at $(-3,-1)$ and $(5,11)$. Center at $(1,5)$, width $=8$, height $=12$.

• For $B$: divide the inequality by $144$:

  $$\frac{(\alpha -2{)}^2}{9}+\frac{(\beta -6{)}^2}{16}\le 1.$$

  So $B$ is an ellipse centered at $(2,6)$ with semi-axes $a=3$ in the $\alpha$-direction and $b=4$ in the $\beta$-direction. Its $x$-range (α-range) is $2\pm3=[-1,5]$; its $y$-range (β-range) is $6\pm4=[2,10]$

Step 2 — Compare ranges

• Rectangle $A$ spans $\alpha\in[-3,5]$ and $\beta\in[-1,11]$.

• Ellipse $B$ spans $\alpha\in[-1,5]$ and $\beta\in[2,10]$.

Observe that every $\alpha$-value of $B$ lies inside the $\alpha$-range of $A$ $(-1$ is between $-3$ and $5)$, and every $\beta$-value of $B$ lies inside the $\beta$-range of $A$ ($2$ to $10$ is inside $-1$ to $11$). Because the ellipse is entirely contained within those coordinate bounds, the ellipse $B$ lies completely inside the rectangle $A$.

Step 3 — Give set relationships

• $B\subseteq A$.

• The containment is proper because there exist points of $A$ not in $B$. Example: take $(-3,5)\in \mathrm{A\; (left\; edge\; of\; the\; rectangle).\; Test\; in }$$B$:
  $16(-3-2)^2+9(5-6)^2=16\cdot25+9\cdot1=400+9=409>144$. So $(-3,5)\notin B$. Thus $B\subsetneq A$.

• Therefore $A\cap B = B$ and $A\cup B = A$. Also $A\setminus B$ is non-empty (points of the rectangle outside the ellipse).

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🖼️ Image Solution

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✅ Conclusion & Video Solution

Final relationships:

$$\boxed{{\displaystyle B⊊A,A\cap B=B,A\cup B=A,A\setminus B\mathrm{\ne }\mathrm{\varnothing }\mathrm{.}}}$$

In words: the ellipse $B$ lies entirely inside the rectangle $A$, but the rectangle has extra area outside the ellipse.

▶️ Watch the full video walkthrough: