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Step-by-Step Solution
Step 1: Identify the given data
The eccentricity of the ellipse is given as $e = \tfrac{1}{2}$, and one of its directrices is $x = 4$.
Step 2: Use the directrix property to find $a$
For an ellipse centered at the origin with its major axis along the x-axis, the directrix is given by $x = \pm \tfrac{a}{e}$. Here, we have:
$$
\frac{a}{e} = 4 \quad \Longrightarrow \quad a = 4e.
$$
Substituting $e = \tfrac{1}{2}$, we get:
$$
a = 4 \times \frac{1}{2} = 2.
$$
Step 3: Calculate $b$ using the relation between $a$, $b$, and $e$
The eccentricity $e$ of an ellipse satisfies $e^2 = 1 - \tfrac{b^2}{a^2}$. Thus,
$$
\left(\frac{1}{2}\right)^2 = 1 - \frac{b^2}{2^2},
$$
which implies
$$
\frac{1}{4} = 1 - \frac{b^2}{4}.
$$
Rearranging,
$$
\frac{b^2}{4} = 1 - \frac{1}{4} = \frac{3}{4}.
$$
Hence,
$$
b^2 = 3 \quad \Longrightarrow \quad b = \sqrt{3}.
$$
Step 4: Write the standard equation of the ellipse
For an ellipse centered at the origin with semi-major axis $a$ along the x-axis and semi-minor axis $b$ along the y-axis, the equation is:
$$
\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1.
$$
Substituting $a = 2$ and $b = \sqrt{3}$, we get
$$
\frac{x^2}{4} + \frac{y^2}{3} = 1.
$$
Step 5: Simplify to get the final form
Multiplying throughout by 12 to clear the denominators:
$$
3x^2 + 4y^2 = 12.
$$
This is the required equation of the ellipse.
