© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Recall the standard definition of an ellipse
An ellipse is the set of points for which the ratio of the distance from a focus to the distance from the corresponding directrix is a constant called the eccentricity, denoted by $e$. If $F$ is a focus and $D$ is the directrix, then any point $P$ on the ellipse satisfies
$PF = e \times PD.$
Step 2: Identify the given information
One focus is at the origin $(0,0)$.
The directrix is the vertical line $x = 4.$
The eccentricity is $e = \frac{1}{2}.$
We want to find the semi-major axis length $a.$
Step 3: Use the known form of directrix for a shifted ellipse
For an ellipse that has its center at the origin (before shifting) and major axis along the $x$-axis, a focus is at $(c,0)$ with $c = a e$, and its corresponding directrix is given by $x = \frac{a}{e}.$
However, in our problem, the ellipse is shifted so that this focus $(c,0)$ is moved to the origin $(0,0)$. When the ellipse is shifted left by $c$ units, the directrix also shifts left by the same amount $c$.
Step 4: Express the directrix in terms of $a$ and $e$ after shifting
If initially the directrix is $x = \frac{a}{e}$ (centered ellipse) and we shift left by $c = a e$, the new directrix will be
$x = \frac{a}{e} - a e = a\Bigl(\frac{1}{e} - e\Bigr).$
We are given that, after shifting, the directrix is $x = 4.$ Hence,
$4 = a\Bigl(\frac{1}{e} - e\Bigr).$
Step 5: Substitute $e = \tfrac{1}{2}$ and solve for $a$
With $e = \tfrac{1}{2}$, compute $\bigl(\tfrac{1}{e} - e\bigr)$:
$\frac{1}{e} - e = 2 - \frac{1}{2} = \frac{3}{2}.$
So,
$4 = a \times \frac{3}{2} \quad \Longrightarrow \quad a = \frac{4 \times 2}{3} = \frac{8}{3}.$
Step 6: Conclude the semi-major axis length
Thus, the length of the semi-major axis of the given ellipse is
$\boxed{\frac{8}{3}}.$
