The eccentricity of the ellipse 25x^2 + 16y^2 | vedclass

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(C) The given equation is $25x^2 + 16y^2 = 100$.Dividing both sides by $100$,we get $\frac{25x^2}{100} + \frac{16y^2}{100} = 1$,which simplifies to $\

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$\frac{3}{5}$

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(C) The given equation is $25x^2 + 16y^2 = 100$.Dividing both sides by $100$,we get $\frac{25x^2}{100} + \frac{16y^2}{100} = 1$,which simplifies to $\frac{x^2}{4} + \frac{y^2}{6.25} = 1$.Comparing this with the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,we have $a^2 = 4$ and $b^2 = \frac{100}{16} = \frac{25}{4} = 6.25$.Since $b^2 > a^2$,the ellipse is vertical,and the eccentricity $e$ is given by $a^2 = b^2(1 - e^2)$.Substituting the values: $4 = \frac{25}{4}(1 - e^2)$.$1 - e^2 = \frac{16}{25}$.$e^2 = 1 - \frac{16}{25} = \frac{9}{25}$.Therefore,$e = \sqrt{\frac{9}{25}} = \frac{3}{5}$.

The eccentricity of the ellipse $25x^2 + 16y^2 = 100$ is

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