For the ellipse frac{x^2}{64} + frac{y^2}{28} | vedclass

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Question

(A) The given equation of the ellipse is $\frac{x^2}{64} + \frac{y^2}{28} = 1$.Comparing this with the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2

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$\frac{3}{4}$

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(A) The given equation of the ellipse is $\frac{x^2}{64} + \frac{y^2}{28} = 1$.Comparing this with the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,we have $a^2 = 64$ and $b^2 = 28$.Since $a^2 > b^2$,the eccentricity $e$ is given by the formula $e^2 = 1 - \frac{b^2}{a^2}$.Substituting the values,we get $e^2 = 1 - \frac{28}{64}$.$e^2 = \frac{64 - 28}{64} = \frac{36}{64}$.Taking the square root,$e = \sqrt{\frac{36}{64}} = \frac{6}{8} = \frac{3}{4}$.

For the ellipse $\frac{x^2}{64} + \frac{y^2}{28} = 1$,the eccentricity is

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