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

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(A) Given equation of the ellipse: $x^2+4 y^2+2 x+16 y+13=0$Completing the square for $x$ and $y$ terms:$(x^2+2x+1) + 4(y^2+4y+4) + 13 - 1 - 16 = 0$$(

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

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(A) Given equation of the ellipse: $x^2+4 y^2+2 x+16 y+13=0$Completing the square for $x$ and $y$ terms:$(x^2+2x+1) + 4(y^2+4y+4) + 13 - 1 - 16 = 0$$(x+1)^2 + 4(y+2)^2 = 4$Dividing by $4$:$\frac{(x+1)^2}{4} + \frac{(y+2)^2}{1} = 1$Comparing with the standard form $\frac{X^2}{a^2} + \frac{Y^2}{b^2} = 1$,we get $a^2 = 4$ and $b^2 = 1$.Since $a^2 > b^2$,the eccentricity $e$ is given by $b^2 = a^2(1 - e^2)$.$1 = 4(1 - e^2)$$1 - e^2 = \frac{1}{4}$$e^2 = 1 - \frac{1}{4} = \frac{3}{4}$$e = \frac{\sqrt{3}}{2}$

The eccentricity of the ellipse $x^2+4 y^2+2 x+16 y+13=0$ is

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