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

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(A) Given the equation of the ellipse: $25x^2 + 16y^2 = 400$. Dividing both sides by $400$,we get: $\frac{x^2}{16} + \frac{y^2}{25} = 1$. Here,$a^2 =

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$3/5$

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(A) Given the equation of the ellipse: $25x^2 + 16y^2 = 400$. Dividing both sides by $400$,we get: $\frac{x^2}{16} + \frac{y^2}{25} = 1$. Here,$a^2 = 25$ and $b^2 = 16$. Since $a^2 > b^2$ is not the case here (as $25 > 16$),the major axis is along the $y$-axis. For an ellipse $\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$,the eccentricity $e$ is given by $e = \sqrt{1 - \frac{b^2}{a^2}}$. Substituting the values: $e = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{25 - 16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5}$.

The eccentricity of the ellipse $25x^2 + 16y^2 = 400$ is

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