The eccentricity of the ellipse x^2+4y^2+2x+1 | vedclass

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(A) Given the equation of the ellipse: $x^2+4y^2+2x+16y+13=0$. Rearranging the terms: $(x^2+2x) + 4(y^2+4y) + 13 = 0$. Completing the squares: $(x^2+2

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Given the equation of the ellipse: $x^2+4y^2+2x+16y+13=0$. Rearranging the terms: $(x^2+2x) + 4(y^2+4y) + 13 = 0$. Completing the squares: $(x^2+2x+1) + 4(y^2+4y+4) + 13 - 1 - 16 = 0$. $(x+1)^2 + 4(y+2)^2 = 4$. 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 $b^2 = a^2(1 - e^2)$. $1 = 4(1 - e^2) \Rightarrow 1 = 4 - 4e^2$. $4e^2 = 3 \Rightarrow e^2 = \frac{3}{4}$. Thus,$e = \frac{\sqrt{3}}{2}$.

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

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