If x ^ 2 + 9 y ^ 2 - 4x + 3 = 0, x, y ∈ R, then x and y...

The Given equation is ${x}^{2} + 9{y}^{2} - 4x + 3 = 0$

$⇒ 9{y}^{2} + 0y + \left({x}^{2} - 4x + 3\right) = 0$

Make quadratic of $y$, we have $D \ge 0$ As it gives real values

$⇒ 0 - 4 \times 9 \times \left({x}^{2} - 4x + 3\right) \ge 0$

$⇒ {x}^{2} - 3x - x + 3 \le 0$

$⇒ \left(x - 3\right)\left(x - 1\right) \le 0$

$x \in \left[1, 3\right]$

Now making quadratic in $x$ equation is ${x}^{2} - 4x + 3 + 9{y}^{2} = 0$

$D \ge 0$

$16 - 4 \times \left(3 + 9{y}^{2}\right) \ge 0$

$⇒ 4 - 3 - 9{y}^{2} \ge 0$

$⇒ 9{y}^{2} \le 1$

$⇒ y \in \left[\dfrac{- 1}{3}, \dfrac{1}{3}\right]$