# The eccentricity of an ellipse , with its centre at origin is $\large\frac{1}{2}$, If one of the directrices is $x=4$ then the equation of the ellipse is

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

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.