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

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(B) The given equation of the ellipse is $9x^2 + 5y^2 - 18x - 20y - 16 = 0$.Rearranging the terms,we get $9(x^2 - 2x) + 5(y^2 - 4y) = 16$.Completing t

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$\frac{2}{3}$

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(B) The given equation of the ellipse is $9x^2 + 5y^2 - 18x - 20y - 16 = 0$.Rearranging the terms,we get $9(x^2 - 2x) + 5(y^2 - 4y) = 16$.Completing the square,we have $9(x^2 - 2x + 1) + 5(y^2 - 4y + 4) = 16 + 9 + 20$.This simplifies to $9(x - 1)^2 + 5(y - 2)^2 = 45$.Dividing by $45$,we obtain the standard form $\frac{(x - 1)^2}{5} + \frac{(y - 2)^2}{9} = 1$.Here,$a^2 = 5$ and $b^2 = 9$. Since $b^2 > a^2$,the ellipse is vertical.The eccentricity $e$ is given by $e = \sqrt{1 - \frac{a^2}{b^2}} = \sqrt{1 - \frac{5}{9}} = \sqrt{\frac{4}{9}} = \frac{2}{3}$.

The eccentricity of the ellipse $9x^2 + 5y^2 - 18x - 20y - 16 = 0$ is:

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