The eccentricity of the ellipse 4x^2 + 9y^2 + | vedclass

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(D) Given equation: $4x^2 + 9y^2 + 8x + 36y + 4 = 0$Rearranging the terms: $4(x^2 + 2x) + 9(y^2 + 4y) = -4$Completing the square: $4(x^2 + 2x + 1) + 9

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$\frac{\sqrt{5}}{3}$

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(D) Given equation: $4x^2 + 9y^2 + 8x + 36y + 4 = 0$Rearranging the terms: $4(x^2 + 2x) + 9(y^2 + 4y) = -4$Completing the square: $4(x^2 + 2x + 1) + 9(y^2 + 4y + 4) = -4 + 4 + 36$$4(x + 1)^2 + 9(y + 2)^2 = 36$Dividing by $36$: $\frac{(x + 1)^2}{9} + \frac{(y + 2)^2}{4} = 1$Here,$a^2 = 9$ and $b^2 = 4$. Since $a^2 > b^2$,the eccentricity $e$ is given by $e = \sqrt{1 - \frac{b^2}{a^2}}$.$e = \sqrt{1 - \frac{4}{9}} = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3}$.

The eccentricity of the ellipse $4x^2 + 9y^2 + 8x + 36y + 4 = 0$ is

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