What is the equation of the ellipse with foci | vedclass

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(A) The foci are given by $(\pm ae, 0) = (\pm 2, 0)$,so $ae = 2$.Given $e = \frac{1}{2}$,we have $a(\frac{1}{2}) = 2$,which implies $a = 4$.Using the

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$3x^2 + 4y^2 = 48$

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(A) The foci are given by $(\pm ae, 0) = (\pm 2, 0)$,so $ae = 2$.Given $e = \frac{1}{2}$,we have $a(\frac{1}{2}) = 2$,which implies $a = 4$.Using the relation $b^2 = a^2(1 - e^2)$,we get $b^2 = 16(1 - \frac{1}{4}) = 16(\frac{3}{4}) = 12$.The equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.Substituting the values,we get $\frac{x^2}{16} + \frac{y^2}{12} = 1$.Multiplying by $48$,we obtain $3x^2 + 4y^2 = 48$.

What is the equation of the ellipse with foci $(\pm 2, 0)$ and eccentricity $e = \frac{1}{2}$?

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