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

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

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

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

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

$\Rightarrow x^2-3x-x+3\le 0$

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

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

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

$D\ge 0$

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

$\Rightarrow 4-3-9y^2\ge 0$

$\Rightarrow 9y^2\le 1$

$\Rightarrow y\in \left[\frac{-1}{3},\frac{1}{3}\right]$