A focus of an ellipse is at the origin. The directrix is the line $$x = 4$$ and the eccentricity is $$1/2$$. Then the length of the semi-major axis is

For any point $$P(x,y)$$ on an ellipse, the ratio of its distance from a focus to its distance from the corresponding directrix is the eccentricity $$e$$.

Here

• focus : $$S(0,0)$$ (the origin)
• directrix : $$x = 4$$ (vertical line)
• eccentricity : $$e = \dfrac12$$

Using the definition, we write

$$\dfrac{\text{distance } PS}{\text{distance of } P \text{ from } x = 4} \;=\; \dfrac12$$

Distance $$PS = \sqrt{x^{2}+y^{2}}$$
Distance of $$P(x,y)$$ from the line $$x=4$$ is $$|x-4|$$.

Hence

$$\sqrt{x^{2}+y^{2}} \;=\; \dfrac12\,|x-4|$$

Squaring both sides:

$$x^{2}+y^{2} \;=\; \dfrac14\,(x-4)^{2}$$

Multiply by 4 to clear the denominator:

$$4x^{2}+4y^{2} \;=\; (x-4)^{2}$$

Expand and rearrange:

$$(x-4)^{2}=x^{2}-8x+16$$
$$4x^{2}+4y^{2}-x^{2}+8x-16=0$$
$$3x^{2}+4y^{2}+8x-16=0$$

Complete the square in $$x$$:

$$3\bigl(x^{2}+\tfrac83x\bigr)+4y^{2}-16=0$$
Inside the brackets: $$x^{2}+\tfrac83x = \bigl(x+\tfrac43\bigr)^{2}-\bigl(\tfrac43\bigr)^{2}$$
$$= \bigl(x+\tfrac43\bigr)^{2}-\tfrac{16}{9}$$

Substitute back:

$$3\Bigl[\bigl(x+\tfrac43\bigr)^{2}-\tfrac{16}{9}\Bigr]+4y^{2}-16 = 0$$
$$3\bigl(x+\tfrac43\bigr)^{2}-\tfrac{16}{3}+4y^{2}-16=0$$
$$3\bigl(x+\tfrac43\bigr)^{2}+4y^{2} = \tfrac{16}{3}+16 = \tfrac{64}{3}$$

Divide the whole equation by $$\dfrac{64}{3}$$ to get the standard form:

$$\dfrac{3}{64/3}\bigl(x+\tfrac43\bigr)^{2} + \dfrac{4}{64/3}\,y^{2} = 1$$

Simplify the coefficients:

$$\dfrac{9}{64}\bigl(x+\tfrac43\bigr)^{2} + \dfrac{3}{16}\,y^{2} = 1$$

Rewrite each coefficient as the reciprocal of a square:

$$\dfrac{\bigl(x+\tfrac43\bigr)^{2}}{64/9} + \dfrac{y^{2}}{16/3} = 1$$

The larger denominator $$\dfrac{64}{9}$$ corresponds to $$a^{2}$$, the square of the semi-major axis.

Therefore

$$a = \sqrt{\dfrac{64}{9}} = \dfrac{8}{3}$$

Option A which is: $$\dfrac{8}{3}$$