The eccentricity of the ellipse frac{(x - 1)^ | vedclass

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(A) Given the equation of the ellipse: $\frac{(x - 1)^2}{9} + \frac{(y + 1)^2}{25} = 1$.Comparing this with the standard form $\frac{(x - h)^2}{b^2} +

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$4/5$

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(A) Given the equation of the ellipse: $\frac{(x - 1)^2}{9} + \frac{(y + 1)^2}{25} = 1$.Comparing this with the standard form $\frac{(x - h)^2}{b^2} + \frac{(y - k)^2}{a^2} = 1$,we have $b^2 = 9$ and $a^2 = 25$.Since $a^2 > b^2$,the major axis is vertical.The formula for eccentricity $e$ is $b^2 = a^2(1 - e^2)$.Substituting the values: $9 = 25(1 - e^2)$.$1 - e^2 = \frac{9}{25}$.$e^2 = 1 - \frac{9}{25} = \frac{16}{25}$.$e = \sqrt{\frac{16}{25}} = \frac{4}{5}$.

The eccentricity of the ellipse $\frac{(x - 1)^2}{9} + \frac{(y + 1)^2}{25} = 1$ is

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