1. Understanding the Ellipse and Focal Distances

For an ellipse defined by the standard equation $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ where $a > b$, the semi-major axis is $a$ and the semi-minor axis is $b$. The foci are located at $S_1(-ae, 0)$ and $S_2(ae, 0)$, where $e$ is the eccentricity of the ellipse.

A fundamental property of an ellipse is that for any point $P(x_1, y_1)$ on the ellipse, the sum of its focal distances is constant and equal to $2a$. That is, $PS_1 + PS_2 = 2a$. The individual focal distances are given by: $PS_1 = a - ex_1$ $PS_2 = a + ex_1$ These formulas are crucial for solving problems involving focal distances. The product of these focal distances is: $PS_1 \cdot PS_2 = (a - ex_1)(a + ex_1) = a^2 - e^2 x_1^2$

We are also given the relationship between $a$, $b$, and $e$: $b^2 = a^2(1 - e^2)$ This equation links the dimensions of the ellipse to its eccentricity. The eccentricity $e$ for an ellipse always satisfies $0 < e < 1$.

2. Setting Up the Equations

We are given a point $P\left(\sqrt{3}, \frac{1}{2}\right)$ on the ellipse and the product of its focal distances as $\frac{7}{4}$. Let $x_1 = \sqrt{3}$ and $y_1 = \frac{1}{2}$.

Using the product of focal distances formula: $a^2 - e^2 x_1^2 = \frac{7}{4}$ Substitute $x_1 = \sqrt{3}$: $a^2 - e^2 (\sqrt{3})^2 = \frac{7}{4}$ $a^2 - 3e^2 = \frac{7}{4} \quad (*)$ Explanation: This equation directly uses the given information about the product of focal distances and relates $a^2$ and $e^2$. Our goal is to find $e$, so we need to express $a^2$ in terms of $e^2$ or vice versa.

Since the point $P\left(\sqrt{3}, \frac{1}{2}\right)$ lies on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, it must satisfy the ellipse's equation: $\frac{(\sqrt{3})^2}{a^2} + \frac{\left(\frac{1}{2}\right)^2}{b^2} = 1$ $\frac{3}{a^2} + \frac{\frac{1}{4}}{b^2} = 1$ $\frac{3}{a^2} + \frac{1}{4b^2} = 1 \quad (**)$ Explanation: This step utilizes the fact that the given point is on the ellipse, providing another equation involving $a^2$ and $b^2$.

3. Expressing $a^2$ in terms of $e^2$

From equation $(*)$, we can express $a^2$ in terms of $e^2$: $a^2 = \frac{7}{4} + 3e^2 \quad (1)$ Explanation: We isolate $a^2$ to substitute it into other equations, eventually leading to an equation solely in terms of $e^2$.

Next, we use the relationship $b^2 = a^2(1 - e^2)$ to eliminate $b^2$ from equation $(**)$: $\frac{3}{a^2} + \frac{1}{4a^2(1-e^2)} = 1 \quad (2)$ Explanation: We replace $b^2$ with $a^2(1-e^2)$ because we want to reduce the number of variables. Our ultimate goal is to solve for $e$.

4. Substituting into the Ellipse Equation and Solving for $e^2$

Now, substitute the expression for $a^2$ from equation (1) into equation (2): $\frac{3}{\left(\frac{7}{4} + 3e^2\right)} + \frac{1}{4\left(\frac{7}{4} + 3e^2\right)(1-e^2)} = 1$ Explanation: This is the crucial substitution that transforms the equation into one involving only $e^2$. This is a common strategy in coordinate geometry problems: use all given conditions to form a system of equations and then eliminate variables until you have an equation for the desired unknown.

To simplify, let's multiply the entire equation by $4\left(\frac{7}{4} + 3e^2\right)(1-e^2)$ to clear the denominators: $3 \cdot 4(1-e^2) + 1 = 4\left(\frac{7}{4} + 3e^2\right)(1-e^2)$ $12(1-e^2) + 1 = (7 + 12e^2)(1-e^2)$ $12 - 12e^2 + 1 = 7 - 7e^2 + 12e^2 - 12e^4$ $13 - 12e^2 = 7 + 5e^2 - 12e^4$ Rearrange the terms to form a quadratic equation in $e^2$: $12e^4 - 17e^2 + 6 = 0$ Explanation: We expanded and rearranged the terms to get a standard form of a quadratic equation. This equation is quadratic in $e^2$, which means we can solve for $e^2$ using factorization or the quadratic formula.

Let $E = e^2$. The equation becomes: $12E^2 - 17E + 6 = 0$ We can factor this quadratic equation. We look for two numbers that multiply to $12 \times 6 = 72$ and add up to $-17$. These numbers are $-9$ and $-8$. $12E^2 - 9E - 8E + 6 = 0$ $3E(4E - 3) - 2(4E - 3) = 0$ $(3E - 2)(4E - 3) = 0$ This gives two possible values for $E = e^2$: $3E - 2 = 0 \implies E = e^2 = \frac{2}{3}$ $4E - 3 = 0 \implies E = e^2 = \frac{3}{4}$

5. Solving for Eccentricity ($e$)

Now, we find the values of $e$ by taking the square root. Since eccentricity $e$ must be positive for an ellipse:

• From $e^2 = \frac{2}{3}$: $e_1 = \sqrt{\frac{2}{3}} = \frac{\sqrt{2}}{\sqrt{3}} = \frac{\sqrt{6}}{3}$
• From $e^2 = \frac{3}{4}$: $e_2 = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$

Self-check: Both values $\frac{\sqrt{6}}{3} \approx \frac{2.449}{3} \approx 0.816$ and $\frac{\sqrt{3}}{2} \approx \frac{1.732}{2} \approx 0.866$ are between 0 and 1, which is a necessary condition for the eccentricity of an ellipse. Thus, both are valid eccentricities.

6. Calculating the Absolute Difference

The problem asks for the absolute difference of the eccentricities of two such ellipses. We need to calculate $|e_1 - e_2|$. $|e_1 - e_2| = \left|\frac{\sqrt{6}}{3} - \frac{\sqrt{3}}{2}\right|$ To subtract these, we find a common denominator, which is 6: $|e_1 - e_2| = \left|\frac{2\sqrt{6}}{6} - \frac{3\sqrt{3}}{6}\right| = \left|\frac{2\sqrt{6} - 3\sqrt{3}}{6}\right|$ Now, we need to determine if $2\sqrt{6}$ is greater or smaller than $3\sqrt{3}$. Compare $(2\sqrt{6})^2 = 4 \times 6 = 24$ with $(3\sqrt{3})^2 = 9 \times 3 = 27$. Since $24 < 27$, it means $2\sqrt{6} < 3\sqrt{3}$. Therefore, the expression inside the absolute value is negative. $|e_1 - e_2| = -\left(\frac{2\sqrt{6} - 3\sqrt{3}}{6}\right) = \frac{3\sqrt{3} - 2\sqrt{6}}{6}$ To match the options, we can simplify this expression. Divide both terms in the numerator by $\sqrt{3}$: $= \frac{\sqrt{3}(3 - 2\sqrt{2})}{6} = \frac{3 - 2\sqrt{2}}{2\sqrt{3}}$ This expression matches option (C).

Let me recheck the options and my calculation. Option (A) is $\frac{1-2 \sqrt{2}}{\sqrt{3}}$. Option (B) is $\frac{1-\sqrt{3}}{\sqrt{2}}$. Option (C) is $\frac{3-2 \sqrt{2}}{2 \sqrt{3}}$. Option (D) is $\frac{3-2 \sqrt{2}}{3 \sqrt{2}}$.

Ah, the given correct answer is (A). Let me carefully re-evaluate. Perhaps there was a mistake in my algebraic manipulation or interpretation.

Let's re-examine the options and the result $\frac{3 - 2\sqrt{2}}{2\sqrt{3}}$. This can also be written as $\frac{3\sqrt{3} - 2\sqrt{6}}{6}$. The provided correct answer is (A) $\frac{1-2 \sqrt{2}}{\sqrt{3}}$. This can be written as $\frac{\sqrt{3}-2\sqrt{6}}{3}$. This is quite different from what I got. Let me check the initial steps.

$a^2 - 3e^2 = \frac{7}{4}$ (This is correct) $\frac{3}{a^2} + \frac{1}{4b^2} = 1$ (This is correct) $b^2 = a^2(1-e^2)$ (This is correct) $\frac{3}{a^2} + \frac{1}{4a^2(1-e^2)} = 1$ (This is correct) Substitute $a^2 = \frac{7}{4} + 3e^2$: $\frac{3}{\frac{7}{4} + 3e^2} + \frac{1}{4(\frac{7}{4} + 3e^2)(1-e^2)} = 1$ (This is correct) Let $A = \frac{7}{4} + 3e^2$. $\frac{3}{A} + \frac{1}{4A(1-e^2)} = 1$ Multiply