1. Form of the ellipse

Since the centre is at the origin and one directrix is of the form $x=4$, the major axis must be along the $x$-axis. So the ellipse is of the form

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, \qquad a>b$

with eccentricity

$e=\frac{c}{a}=\frac12.$

2. Directrix of the ellipse

For the ellipse

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,$

the directrices are

$x=\pm \frac{a}{e}.$

Given one directrix is

$x=4,$

so

$\frac{a}{e}=4.$

Since $e=\frac12$,

$\frac{a}{1/2}=4 \implies 2a=4 \implies a=2.$

Hence,

$a^2=4.$

3. Find $b^2$ using eccentricity

For an ellipse,

$b^2=a^2(1-e^2).$

So,

$b^2=4\left(1-\frac14\right)=4\cdot \frac34=3.$

4. Equation of the ellipse

Thus the equation is

$\frac{x^2}{4}+\frac{y^2}{3}=1.$

Multiplying by $12$,

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

5. Check options

• A: $4x^2+3y^2=1$ ❌
• B: $3x^2+4y^2=12$ ✅
• C: $4x^2+3y^2=12$ ❌
• D: $3x^2+4y^2=1$ ❌

Therefore, the correct option is B.