Question

If the equation $x^2+9y^2-4x+3=0$ is satisfied for all real values of $x$ and $y,$ then

• A. $x\in [1,3]$
• B. $y\in [-\frac{1}{3},\frac{1}{3}]$
• C. $x\in (\frac{1}{3},3]$
• D. $y\in (-3,-\frac{1}{3})$

Answer 1

Answer: (A), (B)

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

Given equation is
$x^2+9y^2-4x+3=0...\left(1\right)$
The above equation is quadratic in terms of both variables $x$ and $y.$

Case$1:$ Considering equation $\left(1\right)$ as quadratic equation in terms of $x$
$x^2-4x+(9y^2+3)=0$
For $x\in R\Rightarrow D\ge 0$
$\Rightarrow 16-4(9y^2+3)\ge 0\Rightarrow 9y^2\le 1\Rightarrow y\in [-\frac{1}{3},\frac{1}{3}]$

Case$2:$ Considering equation $\left(1\right)$ as quadratic equation in terms of $y$
$9y^2+0\cdot y+(x^2-4x+3)=0$
For $y\in R\Rightarrow D\ge 0$
$\Rightarrow 0-4\times 9\times (x^2-4x+3)\ge 0\Rightarrow x^2-4x+3\le 0\Rightarrow x\in [1,3]$