For an ellipse with eccentricity e = frac{1}{ | vedclass

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(B) Given that,eccentricity $e = \frac{1}{2}$ and the equation of the directrix is $x = \frac{a}{e} = 4$. From $\frac{a}{e} = 4$,we get $a = 4 	imes

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$3x^2 + 4y^2 = 12$

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(B) Given that,eccentricity $e = \frac{1}{2}$ and the equation of the directrix is $x = \frac{a}{e} = 4$. From $\frac{a}{e} = 4$,we get $a = 4 	imes e = 4 	imes \frac{1}{2} = 2$. Thus,$a^2 = 4$. Using the relation $b^2 = a^2(1 - e^2)$,we have: $b^2 = 4(1 - (\frac{1}{2})^2) = 4(1 - \frac{1}{4}) = 4(\frac{3}{4}) = 3$. Therefore,the equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. Substituting the values,we get $\frac{x^2}{4} + \frac{y^2}{3} = 1$. Multiplying by $12$,we get $3x^2 + 4y^2 = 12$.

For an ellipse with eccentricity $e = \frac{1}{2}$,the centre is at the origin. If one directrix is $x = 4$,then the equation of the ellipse is:

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