If $\mathrm{x}^{2}+9 \mathrm{y}^{2}-4 \mathrm{x}+3=0, \mathrm{x}, \mathrm{y} \in \mathbb{R}$, then $\mathrm{x}$ and $\mathrm{y}$ respectively lie in the intervals:
Correct Option: , 4
$x^{2}+9 y^{2}-4 x+3=0$
$\left(x^{2}-4 x\right)+\left(9 y^{2}\right)+3=0$
$\left(x^{2}-4 x+4\right)+\left(9 y^{2}\right)+3-4=0$
$(x-2)^{2}+(3 y)^{2}=1$
$\frac{(x-2)^{2}}{(1)^{2}}+\frac{y^{2}}{\left(\frac{1}{3}\right)^{2}}=1$ (equation of an ellipse).
As it is equation of an ellipse, $x$ \& $y$ can vary inside the ellipse.
So, $x-2 \in[-1,1]$ and $y \in\left[-\frac{1}{3}, \frac{1}{3}\right]$
$x \in[1,3] y \in\left[-\frac{1}{3}, \frac{1}{3}\right]$
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