The eccentricity of an ellipse whose centre is at the origin is $$\frac{1}{2}$$. If one of its directrices is $$x = -4$$, then the equation of the normal to it at $$\left(1, \frac{3}{2}\right)$$ is:

We have an ellipse whose centre is at the origin, so its standard Cartesian equation can be written as

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

where $$a>b>0$$, the semi-major axis is $$a$$, the semi-minor axis is $$b$$, the focal distance is $$c$$ and the eccentricity is defined by the relation

$$e=\frac{c}{a} \qquad\text{(since for an ellipse }e<1\text{).}$$

For a horizontal ellipse (major axis along the x-axis) the directrices are the two vertical lines

$$x=\frac{a^{2}}{c}\quad\text{and}\quad x=-\frac{a^{2}}{c}.$$

We are informed that one directrix is $$x=-4$$, so we equate

$$-\frac{a^{2}}{c}=-4\;\Longrightarrow\;\frac{a^{2}}{c}=4.$$

The eccentricity is given as $$e=\frac12$$. Using $$e=\dfrac{c}{a}$$ we get

$$\frac{c}{a}=\frac12\;\Longrightarrow\;c=\frac12\,a.$$

Substituting this value of $$c$$ in the directrix relation $$\dfrac{a^{2}}{c}=4$$ gives

$$\frac{a^{2}}{\,\tfrac12 a\,}=4 \;\Longrightarrow\; 2a=4 \;\Longrightarrow\; a=2.$$

Now we find $$c$$ from $$c=\dfrac12 a$$:

$$c=\frac12\,(2)=1.$$

The fundamental relation $$c^{2}=a^{2}-b^{2}$$ gives $$b$$:

$$b^{2}=a^{2}-c^{2}=2^{2}-1^{2}=4-1=3 \;\Longrightarrow\; b=\sqrt3.$$

Hence the explicit equation of the ellipse is

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

We must now find the equation of the normal at the point $$\left(1,\tfrac32\right)$$. First we verify that this point lies on the ellipse:

$$\frac{1^{2}}{4}+\frac{\left(\tfrac32\right)^{2}}{3} =\frac14+\frac{\tfrac94}{3} =\frac14+\frac34 =1,$$

so the point is indeed on the curve.

To get the slope of the tangent we differentiate the equation of the ellipse implicitly with respect to $$x$$. We start from

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

Differentiating term by term and using $$\dfrac{d}{dx}(y^{2})=2y\,\dfrac{dy}{dx}$$ we obtain

$$\frac{2x}{4}+\frac{2y}{3}\,\frac{dy}{dx}=0.$$

Simplifying the coefficients:

$$\frac{x}{2}+\frac{2y}{3}\,\frac{dy}{dx}=0.$$

Solving for $$\dfrac{dy}{dx}$$ (the slope of the tangent, denoted $$m_{\text{tan}}$$):

$$\frac{2y}{3}\,m_{\text{tan}}=-\frac{x}{2} \;\Longrightarrow\; m_{\text{tan}}=-\frac{x}{2}\cdot\frac{3}{2y} \;\Longrightarrow\; m_{\text{tan}}=-\frac{3x}{4y}.$$

At the specific point $$\left(1,\tfrac32\right)$$ we plug in $$x=1,\;y=\tfrac32$$:

$$m_{\text{tan}}=-\frac{3\cdot1}{4\left(\tfrac32\right)} =-\frac3{4\cdot1.5} =-\frac3{6} =-\frac12.$$

The slope of the normal is the negative reciprocal of the slope of the tangent, so

$$m_{\text{norm}}=-\frac1{m_{\text{tan}}}= -\frac1{-\tfrac12}=2.$$

Now we write the equation of the normal line passing through the given point $$\left(1,\tfrac32\right)$$ with slope $$2$$ using the point-slope form:

$$y-\frac32=2\,(x-1).$$

Expanding the right side:

$$y-\frac32 = 2x-2.$$

Adding $$\dfrac32$$ to both sides gives

$$y = 2x-2+\frac32 =2x-\frac12.$$

To match one of the choices, we can clear the fraction by multiplying every term by $$2$$:

$$2y = 4x-1.$$

Rearranging to the standard linear form $$Ax+By=C$$ we obtain

$$4x-2y=1.$$

This exactly matches Option B.

Hence, the correct answer is Option B.