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(C) The given equation is ${x^2} + 2{y^2} - 2x + 3y + 2 = 0$.Completing the square for $x$ and $y$ terms:$({x^2} - 2x + 1) + 2({y^2} + \frac{3}{2}y +

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$1/\sqrt{2}$

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(C) The given equation is ${x^2} + 2{y^2} - 2x + 3y + 2 = 0$.Completing the square for $x$ and $y$ terms:$({x^2} - 2x + 1) + 2({y^2} + \frac{3}{2}y + \frac{9}{16}) = -2 + 1 + \frac{9}{8}$${(x - 1)^2} + 2{(y + \frac{3}{4})^2} = \frac{1}{8}$Dividing by $\frac{1}{8}$,we get:$\frac{{{{(x - 1)}^2}}}{{1/8}} + \frac{{{{(y + 3/4)}^2}}}{{1/16}} = 1$This is an ellipse of the form $\frac{{{{(x - h)}^2}}}{{{a^2}}} + \frac{{{{(y - k)}^2}}}{{{b^2}}} = 1$ where ${a^2} = \frac{1}{8}$ and ${b^2} = \frac{1}{{16}}$.Since ${a^2} > {b^2}$,the eccentricity $e$ is given by ${b^2} = {a^2}(1 - {e^2})$.$\frac{1}{{16}} = \frac{1}{8}(1 - {e^2})$$1 - {e^2} = \frac{1}{2}$${e^2} = \frac{1}{2}$$e = \frac{1}{{\sqrt{2}}}$.

The eccentricity of the curve represented by the equation ${x^2} + 2{y^2} - 2x + 3y + 2 = 0$ is

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