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

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(B) The 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{

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(B) The 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 $b^2 = 9$. Since $a^2 > b^2$,the eccentricity $e$ is given by the formula: $e = \sqrt{1 - \frac{b^2}{a^2}}$ $e = \sqrt{1 - \frac{9}{25}}$ $e = \sqrt{\frac{25 - 9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}$.

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

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