Solve the Following Questions


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:

  1. $\left[-\frac{1}{3}, \frac{1}{3}\right]$ and $\left[-\frac{1}{3}, \frac{1}{3}\right]$

  2. $\left[-\frac{1}{3}, \frac{1}{3}\right]$ and $[1,3]$

  3. $[1,3]$ and $[1,3]$

  4. $[1,3]$ and $\left[-\frac{1}{3}, \frac{1}{3}\right]$

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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