The distance between the foci of the ellipse | vedclass

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(B) The given equation of the ellipse is $3x^2 + 4y^2 = 48$.Dividing both sides by $48$,we get $\frac{x^2}{16} + \frac{y^2}{12} = 1$.Comparing this wi

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$4$

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(B) The given equation of the ellipse is $3x^2 + 4y^2 = 48$.Dividing both sides by $48$,we get $\frac{x^2}{16} + \frac{y^2}{12} = 1$.Comparing this with the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,we have $a^2 = 16$ and $b^2 = 12$.Thus,$a = 4$ and $b = \sqrt{12} = 2\sqrt{3}$.The eccentricity $e$ is given by $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{12}{16}} = \sqrt{1 - \frac{3}{4}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.The distance between the foci is $2ae = 2 	imes 4 	imes \frac{1}{2} = 4$.

The distance between the foci of the ellipse $3x^2 + 4y^2 = 48$ is

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