$\begin{array}{1 1}(A)\;3x^2+4y^2=1 \\(B)\;3x^2+4y^2=12 \\(C)\;4x^2+3y^2=12 \\(D)\;4x^2+3y^2=1 \end{array}$

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Let the equation of ellipse be $\large\frac{x^2}{a^2}+\frac{y^2}{b^2}$$=1$

Here $ a > b$ because the directrix parallel to y axis.

$b^2=a^2 (1-e^2)$ with $e =\large\frac{1}{2}$

=> $b^2= \large\frac{3}{4} $$a^2$

But $\large \frac{a}{e}$$=4$

=> $a=2$

Required ellipse is $ \large\frac{x^2}{4}+\frac{y^2}{3}$$=1$

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

Hence B is the correct answer.

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