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

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(B) Given equation: $4x^2 + y^2 - 8x + 2y + 1 = 0$Rearranging the terms: $4(x^2 - 2x) + (y^2 + 2y) = -1$Completing the square: $4(x^2 - 2x + 1) + (y^2

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$\sqrt{3}/2$

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(B) Given equation: $4x^2 + y^2 - 8x + 2y + 1 = 0$Rearranging the terms: $4(x^2 - 2x) + (y^2 + 2y) = -1$Completing the square: $4(x^2 - 2x + 1) + (y^2 + 2y + 1) = -1 + 4 + 1$$4(x - 1)^2 + (y + 1)^2 = 4$Dividing by $4$: $\frac{(x - 1)^2}{1} + \frac{(y + 1)^2}{4} = 1$Here,$a^2 = 4$ and $b^2 = 1$. Since $a^2 > b^2$,the ellipse is vertical.The eccentricity $e$ is given by $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$.

The eccentricity of the ellipse $4x^2 + y^2 - 8x + 2y + 1 = 0$ is

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