If x^{2}+9 y^{2}-4 x+3=0,where x, y in R,then | vedclass

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(D) Given equation: $x^{2}+9 y^{2}-4 x+3=0$Rearranging the terms: $(x^{2}-4 x)+(9 y^{2})+3=0$Completing the square for $x$: $(x^{2}-4 x+4)+9 y^{2}+3-4

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

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(D) Given equation: $x^{2}+9 y^{2}-4 x+3=0$Rearranging the terms: $(x^{2}-4 x)+(9 y^{2})+3=0$Completing the square for $x$: $(x^{2}-4 x+4)+9 y^{2}+3-4=0$$(x-2)^{2}+(3 y)^{2}=1$This is the equation of an ellipse: $\frac{(x-2)^{2}}{1^{2}}+\frac{y^{2}}{(1/3)^{2}}=1$For the ellipse $\frac{(x-h)^{2}}{a^{2}}+\frac{(y-k)^{2}}{b^{2}}=1$,the range of $x$ is $[h-a, h+a]$ and the range of $y$ is $[k-b, k+b]$.Here,$h=2, a=1, k=0, b=1/3$.Therefore,$x \in [2-1, 2+1] = [1, 3]$ and $y \in [0-1/3, 0+1/3] = [-\frac{1}{3}, \frac{1}{3}]$.Thus,$x$ and $y$ lie in $[1, 3]$ and $[-\frac{1}{3}, \frac{1}{3}]$ respectively.

If $x^{2}+9 y^{2}-4 x+3=0$,where $x, y \in R$,then $x$ and $y$ respectively lie in the intervals:

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