The Given equation is x2+9y2−4x+3=0
⇒9y2+0y+(x2−4x+3)=0
Make quadratic of y, we have D≥0 As it gives real values
⇒0−4×9×(x2−4x+3)≥0
⇒x2−3x−x+3≤0
⇒(x−3)(x−1)≤0
x∈[1,3]
Now making quadratic in x equation is x2−4x+3+9y2=0
D≥0
16−4×(3+9y2)≥0
⇒4−3−9y2≥0
⇒9y2≤1
⇒y∈[3−1,31]