If x 2 + 9 y 2 - 4 x + 3 = 0 , x , y ∈ R , then x and y respectively lie in the intervals

Step-by-step solution

The Given equation is x 2 + 9 y 2 - 4 x + 3 = 0 ⇒ 9 y 2 + 0 y + x 2 - 4 x + 3 = 0 Make quadratic of y , we have D ≥ 0 As it gives real values ⇒ 0 - 4 × 9 × x 2 - 4 x + 3 ≥ 0 ⇒ x 2 - 3 x - x + 3 ≤ 0 ⇒ x - 3 x - 1 ≤ 0 x ∈ 1 , 3 Now making quadratic in x equation is x 2 - 4 x + 3 + 9 y 2 = 0 D ≥ 0 16 - 4 × 3 + 9 y 2 ≥ 0 ⇒ 4 - 3 - 9 y 2 ≥ 0 ⇒ 9 y 2 ≤ 1 ⇒ y ∈ - 1 3 , 1 3