The eccentricity of the ellipse 9x^2 + 5y^2 - | vedclass

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(B) Given the equation of the ellipse: $9x^2 + 5y^2 - 30y = 0$. Completing the square for the $y$ terms: $9x^2 + 5(y^2 - 6y) = 0$. $9x^2 + 5(y^2 - 6y

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$2/3$

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(B) Given the equation of the ellipse: $9x^2 + 5y^2 - 30y = 0$. Completing the square for the $y$ terms: $9x^2 + 5(y^2 - 6y) = 0$. $9x^2 + 5(y^2 - 6y + 9) = 5(9)$. $9x^2 + 5(y - 3)^2 = 45$. Dividing by $45$,we get: $\frac{x^2}{5} + \frac{(y - 3)^2}{9} = 1$. Here,$a^2 = 5$ and $b^2 = 9$. Since $b^2 > a^2$,the major axis is along the $y$-axis. The eccentricity $e$ is given by $e = \sqrt{1 - \frac{a^2}{b^2}}$. $e = \sqrt{1 - \frac{5}{9}} = \sqrt{\frac{4}{9}} = \frac{2}{3}$.

The eccentricity of the ellipse $9x^2 + 5y^2 - 30y = 0$ is

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