The eccentricity of the ellipse 4x^2 + 25y^2 | vedclass

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Question

(A) Given equation of the ellipse is $4x^2 + 25y^2 = 100$. Dividing both sides by $100$,we get $\frac{4x^2}{100} + \frac{25y^2}{100} = \frac{100}{100}

Vedclass · Accepted Answer

$\frac{\sqrt{21}}{5}$

Vedclass · Answer

(A) Given equation of the ellipse is $4x^2 + 25y^2 = 100$. Dividing both sides by $100$,we get $\frac{4x^2}{100} + \frac{25y^2}{100} = \frac{100}{100}$,which simplifies to $\frac{x^2}{25} + \frac{y^2}{4} = 1$. Comparing this with the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,we have $a^2 = 25$ and $b^2 = 4$. The eccentricity $e$ is given by the formula $e = \sqrt{1 - \frac{b^2}{a^2}}$. Substituting the values,$e = \sqrt{1 - \frac{4}{25}} = \sqrt{\frac{25 - 4}{25}} = \sqrt{\frac{21}{25}} = \frac{\sqrt{21}}{5}$. Thus,the correct option is $A$.

The eccentricity of the ellipse $4x^2 + 25y^2 = 100$ is

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