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જો $x^{2}+9 y^{2}-4 x+3=0, x, y \in R$, હોય તો અનુક્રમે $x$ અને $y$ એ . . . . અંતરાલમાં આવે.
$\left[-\frac{1}{3}, \frac{1}{3}\right]$ અને $\left[-\frac{1}{3}, \frac{1}{3}\right]$
$\left[-\frac{1}{3}, \frac{1}{3}\right]$ અને $[1,3]$
$[1,3]$ અને $[1,3]$
$[1,3]$ અને $\left[-\frac{1}{3}, \frac{1}{3}\right]$
Solution
$x^{2}+9 y^{2}-4 x+3=0$
$\left(x^{2}-4 x\right)+\left(9 y^{2}\right)+3=0$
$\left(x^{2}-4 x+4\right)+\left(9 y^{2}\right)+3-4=0$
$(x-2)^{2}+(3 y)^{2}=1$
$\frac{(x-2)^{2}}{(1)^{2}}+\frac{y^{2}}{\left(\frac{1}{3}\right)^{2}}=1$ (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 $y \in\left[-\frac{1}{3}, \frac{1}{3}\right]$
$x \in[1,3] y \in\left[-\frac{1}{3}, \frac{1}{3}\right]$
