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Ellipse question

2021 · 27 Aug · Shift 1 · Q31
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Ellipse question

2021 · 27 Aug · Shift 1 · Q31

JEE MainMathematicsEllipseMCQ+4 / −1
If x2 + 9y2 −-− 4x + 3 = 0, x, y ∈\in∈ R, then x and y respectively lie in the intervals :
  1. A
    [−13,13]\left[ { - {1 \over 3},{1 \over 3}} \right][−31​,31​] and [−13,13]\left[ { - {1 \over 3},{1 \over 3}} \right][−31​,31​]
  2. B
    [−13,13]\left[ { - {1 \over 3},{1 \over 3}} \right][−31​,31​] and [1, 3]
  3. C
    [1, 3] and [1, 3]
  4. D
    [1, 3] and [−13,13]\left[ { - {1 \over 3},{1 \over 3}} \right][−31​,31​]
View written solutionFree

Correct answer: D

  1. Start with the given equation:

x2+9y2−4x+3=0x^2+9y^2-4x+3=0x2+9y2−4x+3=0

  1. Complete the square in xxx:

x2−4x=(x−2)2−4x^2-4x=(x-2)^2-4x2−4x=(x−2)2−4

So the equation becomes

(x−2)2−4+9y2+3=0(x-2)^2-4+9y^2+3=0(x−2)2−4+9y2+3=0

(x−2)2+9y2−1=0(x-2)^2+9y^2-1=0(x−2)2+9y2−1=0

or

(x−2)2+9y2=1(x-2)^2+9y^2=1(x−2)2+9y2=1

  1. Rewrite in standard ellipse form:

(x−2)21+y21/9=1\frac{(x-2)^2}{1}+\frac{y^2}{1/9}=11(x−2)2​+1/9y2​=1

This is an ellipse centered at (2,0)(2,0)(2,0) with:

  • semi-axis along xxx-direction = 111
  • semi-axis along yyy-direction = 13\frac{1}{3}31​
  1. Hence the possible values of xxx are:

2−1≤x≤2+12-1 \le x \le 2+12−1≤x≤2+1

1≤x≤31 \le x \le 31≤x≤3

So,

x∈[1,3]x \in [1,3]x∈[1,3]

  1. The possible values of yyy are:

−13≤y≤13-\frac{1}{3} \le y \le \frac{1}{3}−31​≤y≤31​

So,

y∈[−13,13]y \in \left[-\frac{1}{3},\frac{1}{3}\right]y∈[−31​,31​]

  1. Compare with the options:
  • A: wrong
  • B: wrong
  • C: wrong
  • D: correct

Therefore, the correct option is D.

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