The correct options are
A x∈[1,3]
B y∈[−13,13]
Given equation is
x2+9y2−4x+3=0 ...(1)
The above equation is quadratic in terms of both variables x and y.
Case 1: Considering equation (1) as quadratic equation in terms of x
x2−4x+(9y2+3)=0
For x∈R⇒D≥0
⇒16−4(9y2+3)≥0⇒9y2≤1⇒y∈[−13,13]
Case 2: Considering equation (1) as quadratic equation in terms of y
9y2+0⋅y+(x2−4x+3)=0
For y∈R⇒D≥0
⇒0−4×9×(x2−4x+3)≥0⇒x2−4x+3≤0⇒x∈[1,3]