JEE Main 2021 · 27 August · morning · Single correct · +4 / −1

If x² + 9y² - 4x + 3 = 0, x, y ∈ R, then x and y respectively lie in the intervals :

  1. (a) [ - 1/3,1/3 ] and [ - 1/3,1/3 ]
  2. (b) [ - 1/3,1/3 ] and [1, 3]
  3. (c) [1, 3] and [1, 3]
  4. (d) [1, 3] and [ - 1/3,1/3 ]
Answer and solution

Answer (d) [1, 3] and [ - 1/3,1/3 ]

Solution

Maths sir · 4 steps

Complete the square in x, and the ellipse tells you exactly how far x and y can stretch.

  1. Complete the square. Group the x terms: x² - 4x = (x-2)² - 4. The equation becomes (x-2)² - 4 + 9y² + 3 = 0, that is (x-2)² + 9y² = 1.

    (x-2)²/1 + y²/(1/3)² = 1

  2. Read off the ranges. This is an ellipse with centre (2, 0), semi-axis 1 along x and 1/3 along y. Both squared terms are non-negative and add to 1, so each is at most 1. That gives (x-2)² ≤ 1 and 9y² ≤ 1.
  3. Solve each. -1 ≤ x - 2 ≤ 1 gives 1 ≤ x ≤ 3. y² ≤ 1/9 gives -1/3 ≤ y ≤ 1/3.
  4. So. x lies in [1, 3] and y lies in [-1/3, 1/3], which is option (d).

The idea to keep. For a real curve written as a sum of squares equal to a constant, each square is at most that constant. Complete the square first so you do not lose the shift of the centre.

Answer from the NTA final answer key. Solution by PrepWiser.

Study Ellipse

Asked about 100 times in JEE Main since 2020. This question is from 2021.