This problem requires us to apply our knowledge of the standard properties of an ellipse, specifically its directrices, eccentricity, and the equation of its normal. Let's break down the concepts and then solve the problem systematically.

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I. Fundamental Concepts and Formulas for an Ellipse

For an ellipse centered at the origin $(0,0)$ with its major axis lying along the x-axis, its standard equation and related properties are as follows:

1. Standard Equation of the Ellipse: $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ Here, $a$ represents the length of the semi-major axis (half the length of the longest diameter), and $b$ represents the length of the semi-minor axis (half the length of the shortest diameter). For an ellipse with the major axis along the x-axis, we always have $a > b$.

2. Eccentricity ($e$): Eccentricity is a crucial parameter that describes the 'roundness' or 'flatness' of an ellipse. It is defined such that $0 < e < 1$. The relationship between the semi-major axis ($a$), semi-minor axis ($b$), and eccentricity ($e$) is given by: $b^2 = a^2(1 - e^2)$ This formula helps connect the dimensions of the ellipse to its shape.

3. Directrices: An ellipse has two directrices, which are lines perpendicular to the major axis. For an ellipse with its major axis along the x-axis and centered at the origin, the equations of its directrices are: $x = \pm \frac{a}{e}$ These lines play a key role in the geometric definition of an ellipse (locus of points where the ratio of distance from a focus to distance from a directrix is constant, equal to $e$).

4. Equation of the Normal to an Ellipse: The normal to a curve at a point is a line perpendicular to the tangent at that point. For an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, the equation of the normal at a point $(x_1, y_1)$ on the ellipse is given by the formula: $\frac{a^2x}{x_1} - \frac{b^2y}{y_1} = a^2 - b^2$ This formula is a direct and efficient way to find the normal's equation without first calculating the tangent's slope.

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II. Step-by-Step Solution

Our strategy is to first determine the complete equation of the ellipse using the given eccentricity and directrix, and then use this information along with the given point to find the equation of the normal.

Step 1: Determine the Equation of the Ellipse

1. Identify Given Information: We are provided with:

   • Eccentricity, $e = \frac{1}{2}$.
   • One of the directrices is $x = -4$.
   • The center of the ellipse is at the origin $(0,0)$.

2. Determine the Orientation of the Major Axis: The given directrix is $x = -4$, which is a vertical line. This immediately tells us that the major axis of the ellipse lies along the x-axis. If the directrix were given as $y = \text{constant}$, the major axis would be along the y-axis, and we would use a different standard equation for the ellipse. This confirms that our chosen standard equation $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is appropriate.

3. Calculate the Semi-Major Axis ($a$): We use the formula for the directrices for an ellipse with its major axis along the x-axis: $x = \pm \frac{a}{e}$. Since the given directrix is $x = -4$, we equate it to the negative form of the formula: $x = -\frac{a}{e}$ Substituting the given values: $-4 = -\frac{a}{1/2}$ $4 = 2a$ $a = 2$ Explanation: This step is crucial. The directrix equation, along with the eccentricity, provides a direct relationship that allows us to determine the length of the semi-major axis, $a$. This value is fundamental as it defines the overall 'size' of the ellipse along its major axis.

4. Calculate the Semi-Minor Axis ($b$): Now that we have $a=2$ and $e=\frac{1}{2}$, we can find $b^2$ using the fundamental relationship between $a, b,$ and $e$: $b^2 = a^2(1 - e^2)$ Substitute the known values: $b^2 = (2)^2 \left(1 - \left(\frac{1}{2}\right)^2\right)$ $b^2 = 4 \left(1 - \frac{1}{4}\right)$ $b^2 = 4 \left(\frac{3}{4}\right)$ $b^2 = 3$ Explanation: This formula connects the lengths of the semi-major and semi-minor axes to the eccentricity. Calculating $b^2$ is essential to fully define the dimensions and shape of the ellipse. We now have $a^2 = 2^2 = 4$ and $b^2 = 3$.

5. Formulate the Equation of the Ellipse: Substitute the calculated values of $a^2=4$ and $b^2=3$ into the standard ellipse equation: $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ $\frac{x^2}{4} + \frac{y^2}{3} = 1$ Explanation: This is the complete equation of the ellipse, uniquely determined by the given eccentricity and directrix.

Step 2: Verify the Given Point on the Ellipse

Before proceeding to find the normal, it is a good practice to verify that the given point $\left(1, \frac{3}{2}\right)$ actually lies on the ellipse we just derived. This acts as a quick check for our calculated ellipse equation and ensures we are working with valid coordinates.

Substitute $x=1$ and $y=\frac{3}{2}$ into the ellipse equation $\frac{x^2}{4} + \frac{y^2}{3} = 1$: $\frac{(1)^2}{4} + \frac{\left(\frac{3}{2}\right)^2}{3}$ $= \frac{1}{4} + \frac{\frac{9}{4}}{3}$ $= \frac{1}{4} + \frac{9}{4 \times 3}$ $= \frac{1}{4} + \frac{9}{12}$ Simplify the second term: $\frac{9}{12} = \frac{3 \times 3}{3 \times 4} = \frac{3}{4}$. $= \frac{1}{4} + \frac{3}{4}$ $= \frac{4}{4} = 1$ Since the left-hand side equals the right-hand side, the point $\left(1, \frac{3}{2}\right)$ indeed lies on the ellipse $\frac{x^2}{4} + \frac{y^2}{3} = 1$. This confirms our ellipse equation is correct.

Step 3: Find the Equation of the Normal

1. Apply the Normal Equation Formula: We use the standard formula for the normal to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at a point $(x_1, y_1)$: $\frac{a^2x}{x_1} - \frac{b^2y}{y_1} = a^2 - b^2$ From our ellipse equation, we have:

   • $a^2 = 4$
   • $b^2 = 3$ The given point is $(x_1, y_1) = \left(1, \frac{3}{2}\right)$.

2. Substitute the Values: Substitute these values into the normal equation formula: $\frac{4x}{1} - \frac{3y}{3/2} = 4 - 3$ $4x - \frac{3y \times 2}{3} = 1$ $4x - 2y = 1$ Explanation: This is the most direct way to find the normal equation using the established formula. The substitution is straightforward, and the subsequent simplification leads directly to the required form.

Step 4: Compare with Options

The derived equation of the normal is $4x - 2y = 1$. Let's compare this with the given options: (A) $2y - x = 2 \implies x - 2y = -2$ (B) $4x - 2y = 1$ (C) $4x + 2y = 7$ (D) $x + 2y = 4$

Our result matches option (B).

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III. Alternative Method (Using Differentiation to find Slope)

For completeness and to reinforce understanding, let's briefly outline how one could find the normal using calculus:

1. Differentiate the ellipse equation implicitly with respect to $x$: Given the ellipse equation $\frac{x^2}{4} + \frac{y^2}{3} = 1$, differentiate both sides: $\frac{2x}{4} + \frac{2y}{3} \frac{dy}{dx} = 0$ $\frac{x}{2} + \frac{2y}{3} \frac{dy}{dx} = 0$
2. Solve for $\frac{dy}{dx}$ (slope of the tangent): $\frac{2y}{3} \frac{dy}{dx} = -\frac{x}{2}$ $\frac{dy}{dx} = -\frac{x}{2} \times \frac{3}{2y}$ $m_T = -\frac{3x}{4y}$
3. Calculate the slope of the tangent ($m_T$) at the point $\left(1, \frac{3}{2}\right)$: $m_T = -\frac{3(1)}{4(3/2)} = -\frac{3}{6} = -\frac{1}{2}$
4. Calculate the slope of the normal ($m_N$): Since the normal is perpendicular to the tangent, their slopes are negative reciprocals: $m_N = -\frac{1}{m_T}$. $m_N = -\frac{1}{(-1/2)} = 2$ 5