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MCQEasyJEE 2025Conic Sections (Parabola, Ellipse, Hyperbola)

JEE Mathematics 2025 Question with Solution

Let A={(α,β)∈R×R:∣α−1∣≤4 and ∣β−5∣≤6}A = \left\{(\alpha, \beta) \in \mathbb{R} \times \mathbb{R} : |\alpha - 1| \leq 4 \text{ and } |\beta - 5| \leq 6\right\} and B={(α,β)∈R×R:16(α−2)2+9(β−6)2≤144}B = \left\{(\alpha, \beta) \in \mathbb{R} \times \mathbb{R} : 16(\alpha - 2)^2 + 9(\beta - 6)^2 \leq 144\right\}. Then:

  • A

    B⊂AB \subset A

  • B

    A∪B={(x,y):−4≤x≤4,−1≤y≤11}A \cup B = \{(x, y) : -4 \leq x \leq 4, -1 \leq y \leq 11 \}

  • C

    neither A⊂BA \subset B nor B⊂AB \subset A

  • D

    A⊂BA \subset B

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: A={(α,β)∈R×R:∣α−1∣≤4 and ∣β−5∣≤6}A = \left\{(\alpha, \beta) \in \mathbb{R} \times \mathbb{R} : |\alpha - 1| \leq 4 \text{ and } |\beta - 5| \leq 6\right\} and B={(α,β)∈R×R:16(α−2)2+9(β−6)2≤144}B = \left\{(\alpha, \beta) \in \mathbb{R} \times \mathbb{R} : 16(\alpha - 2)^2 + 9(\beta - 6)^2 \leq 144\right\}.

Find: the correct relation between the sets AA and BB.

For set AA,

−4≤α−1≤4-4 \leq \alpha - 1 \leq 4

so,

−3≤α≤5-3 \leq \alpha \leq 5

Also,

−6≤β−5≤6-6 \leq \beta - 5 \leq 6

so,

−1≤β≤11-1 \leq \beta \leq 11

Hence, AA is the rectangular region bounded by −3≤α≤5-3 \leq \alpha \leq 5 and −1≤β≤11-1 \leq \beta \leq 11.

For set BB,

16(α−2)2+9(β−6)2≤14416(\alpha - 2)^2 + 9(\beta - 6)^2 \leq 144

Dividing by 144144,

(α−2)29+(β−6)216≤1\frac{(\alpha - 2)^2}{9} + \frac{(\beta - 6)^2}{16} \leq 1

So, BB is an ellipse centered at (2,6)(2, 6) with semi-axis lengths 33 along the α\alpha-direction and 44 along the β\beta-direction.

Therefore, the α\alpha-range of BB is

2−3≤α≤2+32 - 3 \leq \alpha \leq 2 + 3

that is,

−1≤α≤5-1 \leq \alpha \leq 5

And the β\beta-range of BB is

6−4≤β≤6+46 - 4 \leq \beta \leq 6 + 4

that is,

2≤β≤102 \leq \beta \leq 10

These intervals lie completely inside the rectangle ranges of AA:

[−1,5]⊂[−3,5],[2,10]⊂[−1,11][-1, 5] \subset [-3, 5], \qquad [2, 10] \subset [-1, 11]

Hence every point of BB lies in AA.

Therefore, the correct option is A, that is, B⊂AB \subset A.

Shape Interpretation

Given: the set AA is defined by two modulus inequalities and the set BB is defined by a quadratic inequality.

Find: whether one set contains the other.

The inequalities in AA describe a rectangle because each coordinate is restricted independently:

  • ∣α−1∣≤4|\alpha - 1| \leq 4 gives a horizontal strip −3≤α≤5-3 \leq \alpha \leq 5.
  • ∣β−5∣≤6|\beta - 5| \leq 6 gives a vertical strip −1≤β≤11-1 \leq \beta \leq 11.

Their intersection is the rectangle AA.

Now rewrite BB in standard ellipse form:

(α−2)29+(β−6)216≤1\frac{(\alpha - 2)^2}{9} + \frac{(\beta - 6)^2}{16} \leq 1

So the ellipse has center (2,6)(2, 6), horizontal semi-axis 33, and vertical semi-axis 44.

Thus the extreme points of the ellipse occur at:

  • leftmost point: (−1,6)(-1, 6)
  • rightmost point: (5,6)(5, 6)
  • bottommost point: (2,2)(2, 2)
  • topmost point: (2,10)(2, 10)

All these lie inside the rectangle −3≤α≤5-3 \leq \alpha \leq 5 and −1≤β≤11-1 \leq \beta \leq 11. Hence the whole ellipse lies inside the rectangle.

Therefore, B⊂AB \subset A and the correct option is A.

Common mistakes

  • Mistake: taking the semi-axis lengths of the ellipse directly from 1616 and 99. Why it is wrong: the standard form is obtained only after dividing by 144144. What to do instead: first rewrite the ellipse as (α−2)29+(β−6)216≤1\frac{(\alpha - 2)^2}{9} + \frac{(\beta - 6)^2}{16} \leq 1, then read the semi-axes as 33 and 44.

  • Mistake: converting ∣α−1∣≤4|\alpha - 1| \leq 4 into −4≤α≤4-4 \leq \alpha \leq 4. Why it is wrong: the interval must be centered at 11, not at 00. What to do instead: write −4≤α−1≤4-4 \leq \alpha - 1 \leq 4 and then add 11 throughout to get −3≤α≤5-3 \leq \alpha \leq 5.

  • Mistake: deciding set inclusion by comparing only the centers (1,5)(1, 5) and (2,6)(2, 6). Why it is wrong: inclusion depends on the full boundary, not on the centers. What to do instead: compare the complete coordinate ranges or the extreme points of the ellipse with the rectangle.

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