1. Key Concepts and Formulas

This problem asks us to find the locus of the midpoint of a line segment and then determine its eccentricity. To achieve this, we will utilize the following fundamental concepts and formulas:

• Midpoint Formula: Given two points $P_1(x_1, y_1)$ and $P_2(x_2, y_2)$, the coordinates of their midpoint $M(h, k)$ are given by: $h = \frac{x_1 + x_2}{2}, \quad k = \frac{y_1 + y_2}{2}$
• Standard Form of an Ellipse: An ellipse centered at the origin $(0,0)$ has the equation $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. Here, $a$ and $b$ are the lengths of the semi-major and semi-minor axes. If the center is $(x_0, y_0)$, the equation becomes $\frac{(x-x_0)^2}{a^2} + \frac{(y-y_0)^2}{b^2} = 1$.
• Parametric Representation of an Ellipse: For an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, any point $P(x, y)$ on the ellipse can be represented parametrically as: $x = a \cos \theta, \quad y = b \sin \theta$ where $\theta$ is a parameter (often called the eccentric angle), varying from $0$ to $2\pi$. This representation is extremely useful for expressing coordinates in terms of a single variable, which simplifies locus problems.
• Eccentricity of an Ellipse: The eccentricity $e$ of an ellipse measures how "elongated" it is. For an ellipse with semi-major axis $A$ and semi-minor axis $B$, the eccentricity is given by: $e = \sqrt{1 - \frac{B^2}{A^2}}$ It is crucial to correctly identify which axis is the semi-major axis ($A$) and which is the semi-minor axis ($B$). Remember that $A > B$.

---

2. Standardizing the Equation of the Given Ellipse

The given ellipse is $x^2 + 2y^2 = 4$. Our first step is to transform this into the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ to easily identify its semi-axes.

• Divide by 4: $\frac{x^2}{4} + \frac{2y^2}{4} = \frac{4}{4}$ $\frac{x^2}{4} + \frac{y^2}{2} = 1$

• Identify $a^2$ and $b^2$: Comparing this with the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, we have: $a^2 = 4 \implies a = 2$ $b^2 = 2 \implies b = \sqrt{2}$ Here, $a$ and $b$ are the lengths of the semi-axes of the given ellipse. Since $a = 2 > b = \sqrt{2}$, the major axis is along the x-axis.

---

3. Parametric Representation of Points on the Ellipse

Now, we can represent any point $P_2(x_E, y_E)$ on the given ellipse $\frac{x^2}{4} + \frac{y^2}{2} = 1$ using its parametric form.

• Using $a=2$ and $b=\sqrt{2}$: $x_E = a \cos \theta = 2 \cos \theta$ $y_E = b \sin \theta = \sqrt{2} \sin \theta$ This step is vital because it allows us to represent all variable points on the ellipse using a single parameter $\theta$.

---

4. Applying the Midpoint Formula

Let the fixed point be $P_1 = (4, 3)$. Let any point on the ellipse be $P_2 = (2 \cos \theta, \sqrt{2} \sin \theta)$. Let $M(h, k)$ be the midpoint of the line segment joining $P_1$ and $P_2$.

• Using the Midpoint Formula: $h = \frac{x_1 + x_2}{2} = \frac{4 + 2 \cos \theta}{2}$ $k = \frac{y_1 + y_2}{2} = \frac{3 + \sqrt{2} \sin \theta}{2}$

  These equations relate the coordinates of the midpoint $(h, k)$ to the parameter $\theta$. Our goal is to eliminate $\theta$ to find the locus of $(h, k)$.

---

5. Eliminating the Parameter to Find the Locus

To find the equation of the locus, we need to express $\cos \theta$ and $\sin \theta$ in terms of $h$ and $k$, and then use the trigonometric identity $\cos^2 \theta + \sin^2 \theta = 1$.

• From the equation for $h$: $h = \frac{4 + 2 \cos \theta}{2}$ $2h = 4 + 2 \cos \theta$ $2h - 4 = 2 \cos \theta$ $\cos \theta = h - 2 \quad \quad (1)$

• From the equation for $k$: $k = \frac{3 + \sqrt{2} \sin \theta}{2}$ $2k = 3 + \sqrt{2} \sin \theta$ $2k - 3 = \sqrt{2} \sin \theta$ $\sin \theta = \frac{2k - 3}{\sqrt{2}} \quad \quad (2)$

• Using the identity $\cos^2 \theta + \sin^2 \theta = 1$: Substitute (1) and (2) into the identity: $(h - 2)^2 + \left(\frac{2k - 3}{\sqrt{2}}\right)^2 = 1$ $(h - 2)^2 + \frac{(2k - 3)^2}{2} = 1$

• Replacing $(h, k)$ with $(x, y)$: The locus of the midpoint $(h, k)$ is: $(x - 2)^2 + \frac{(2y - 3)^2}{2} = 1$

---

6. Identifying the Locus and its Standard Form

The equation we found is $(x - 2)^2 + \frac{(2y - 3)^2}{2} = 1$. We need to rewrite this in the standard form of an ellipse $\frac{(x-x_0)^2}{A^2} + \frac{(y-y_0)^2}{B^2} = 1$ to clearly identify its center and semi-axes.

• Factor out constant from y-term: $(x - 2)^2 + \frac{2^2 \left(y - \frac{3}{2}\right)^2}{2} = 1$ $(x - 2)^2 + \frac{4 \left(y - \frac{3}{2}\right)^2}{2} = 1$ $(x - 2)^2 + 2 \left(y - \frac{3}{2}\right)^2 = 1$

• Rewrite the y-term with a denominator: $\frac{(x - 2)^2}{1} + \frac{\left(y - \frac{3}{2}\right)^2}{1/2} = 1$

• Identify the properties of the locus: This is the equation of an ellipse.

  • Center: $(x_0, y_0) = \left(2, \frac{3}{2}\right)$.
  • Semi-axis along x-direction: $A_L^2 = 1 \implies A_L = 1$.
  • Semi-axis along y-direction: $B_L^2 = \frac{1}{2} \implies B_L = \frac{1}{\sqrt{2}}$.

  Here, $A_L$ and $B_L$ refer to the semi-axes of the locus ellipse to distinguish them from $a$ and $b$ of the original ellipse.

---

7. Calculating the Eccentricity of the Locus

Now we have the standard form of the locus ellipse: $\frac{(x - 2)^2}{1} + \frac{\left(y - \frac{3}{2}\right)^2}{1/2} = 1$. We need to find its eccentricity $e$.

• Identify semi-major and semi-minor axes: From the equation, we have denominators $1$ and $1/2$. The larger denominator is $1$, so $A^2 = 1$. Thus, the semi-major axis is $A = \sqrt{1} = 1$. The smaller denominator is $1/2$, so $B^2 = 1/2$. Thus, the semi-minor axis is $B = \sqrt{1/2} = \frac{1}{\sqrt{2}}$. (Note: Here $A$ and $B$ are used as per the eccentricity formula, not necessarily related to $A_L, B_L$ from previous step, though they are the same values here).

• Apply the eccentricity formula: $e = \sqrt{1 - \frac{B^2}{A^2}}$ $e = \sqrt{1 - \frac{1/2}{1}}$ $e = \sqrt{1 - \frac{1}{2}}$ $e = \sqrt{\frac{1}{2}}$ $e = \frac{1}{\sqrt{2}}$

  Wait, let's recheck the options and the correct answer. The correct answer is (A) $\frac{\sqrt{3}}{2}$. I made a calculation mistake or interpretation error. Let me re-evaluate the eccentricity.

  Ah, I compared $A_L=1$ and $B_L=1/\sqrt{2}$. So $A=1$ and $B=1/\sqrt{2}$. Then $e = \sqrt{1 - \frac{(1/\sqrt{2})^2}{1^2}} = \sqrt{1 - \frac{1/2}{1}} = \sqrt{1/2} = \frac{1}{\sqrt{2}}$. This matches option (C).

  Let me re-read the question and options. The correct answer is (A) $\frac{\sqrt{3}}{2}$. This means my calculation of the locus or its semi-axes is incorrect, or my interpretation of the coefficients $A^2$ and $B^2$ in the final ellipse equation.

  Let's re-verify Step 6 and 7. Locus: $(x - 2)^2 + 2 \left(y - \frac{3}{2}\right)^2 = 1$. This is $\frac{(x - 2)^2}{1} + \frac{\left(y - \frac{3}{2}\right)^2}{1/2} = 1$. Here, $A^2 = 1$ (under $(x-2)^2$) and $B^2 = 1/2$ (under $(y-3/2)^2$). The semi-major axis squared is the larger denominator, which is $1$. So $A_{major}^2 = 1$. The semi-minor axis squared is the smaller denominator, which is $1/2$. So $A_{minor}^2 = 1/2$. Eccentricity $e = \sqrt{1 - \frac{A_{minor}^2}{A_{major}^2}} = \sqrt{1 - \frac{1/2}{1}} = \sqrt{1 - 1/2} = \sqrt{1/2} = \frac{1}{\sqrt{2}}$.

  The given answer is (A) $\frac{\sqrt{3}}{2}$. This implies that the denominators