The eccentricity of the ellipse y^{2}+4x^{2}- | vedclass

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(A) Given the equation of the ellipse: $y^{2}+4x^{2}-12x+6y+14=0$.Rearranging the terms,we get: $4(x^{2}-3x) + (y^{2}+6y) = -14$.Completing the square

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$\frac{\sqrt{3}}{2}$

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(A) Given the equation of the ellipse: $y^{2}+4x^{2}-12x+6y+14=0$.Rearranging the terms,we get: $4(x^{2}-3x) + (y^{2}+6y) = -14$.Completing the square:$4(x^{2}-3x+\frac{9}{4}) + (y^{2}+6y+9) = -14 + 9 + 9$.$4(x-\frac{3}{2})^{2} + (y+3)^{2} = 4$.Dividing by $4$,we get the standard form: $\frac{(x-\frac{3}{2})^{2}}{1} + \frac{(y+3)^{2}}{4} = 1$.Here,$a^{2}=1$ and $b^{2}=4$. Since $b^{2} > a^{2}$,the major axis is vertical.The eccentricity $e$ is given by $e = \sqrt{1 - \frac{a^{2}}{b^{2}}}$.$e = \sqrt{1 - \frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$.

The eccentricity of the ellipse $y^{2}+4x^{2}-12x+6y+14=0$ is

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