The eccentricity of the ellipse 9x^{2} + 25y^ | vedclass

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(C) Given equation of the ellipse is $9x^{2} + 25y^{2} = 225$. Dividing both sides by $225$,we get: $\frac{9x^{2}}{225} + \frac{25y^{2}}{225} = 1$ $\f

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(C) Given equation of the ellipse is $9x^{2} + 25y^{2} = 225$. Dividing both sides by $225$,we get: $\frac{9x^{2}}{225} + \frac{25y^{2}}{225} = 1$ $\frac{x^{2}}{25} + \frac{y^{2}}{9} = 1$ Comparing this with the standard form $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$,we have $a^{2} = 25$ and $b^{2} = 9$. Thus,$a = 5$ and $b = 3$. The eccentricity $e$ is given by the formula $e = \sqrt{1 - \frac{b^{2}}{a^{2}}}$. $e = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{25 - 9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}$.

The eccentricity of the ellipse $9x^{2} + 25y^{2} = 225$ is

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