Here is a more elaborate, clear, and educational solution to the problem.

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Understanding the Fundamental Properties of an Ellipse

Before we delve into the solution, let's briefly recall the essential definitions and formulas for a standard ellipse. We typically consider an ellipse centered at the origin $(0,0)$ with its major axis along the x-axis. Its equation is given by: $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ where:

• $a$ is the length of the semi-major axis (half the length of the major axis).
• $b$ is the length of the semi-minor axis (half the length of the minor axis).
• By definition for an ellipse, $a > b$.

Key points and formulas relevant to this problem:

1. Vertices: The points where the ellipse intersects its major axis are $(\pm a, 0)$. These are the "endpoints" of the major axis.
2. Foci: The two fixed points inside the ellipse are $(\pm ae, 0)$, where $e$ is the eccentricity.
3. Eccentricity ($e$): A dimensionless parameter that measures how "stretched" or "flattened" an ellipse is. For an ellipse, its value always lies between 0 and 1 (i.e., $0 < e < 1$).
4. Fundamental Relation: The semi-major axis $a$, semi-minor axis $b$, and eccentricity $e$ are related by the equation: $b^2 = a^2(1 - e^2)$ This relation is crucial for connecting the different geometric properties of an ellipse.
5. Length of Latus Rectum (LLR): This is the length of a chord that passes through a focus and is perpendicular to the major axis. Its formula is: $LLR = \frac{2b^2}{a}$

With these foundational concepts in mind, let's approach the problem systematically.

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Step-by-Step Solution

Our goal is to find the eccentricity $e$ using the two pieces of information provided: the length of the latus rectum and the distance between a focus and its nearest vertex.

Step 1: Using the Length of the Latus Rectum (LLR)

The problem states that the length of the latus rectum is 4 units.

• Recall the formula: $LLR = \frac{2b^2}{a}$
• Set up the equation: We are given $LLR = 4$. $\frac{2b^2}{a} = 4$
• Simplify the equation: To make it easier to use later, let's express $b^2$ in terms of $a$. Multiply both sides by $\frac{a}{2}$: $b^2 = 2a \quad \text{(Equation 1)}$ This equation provides a relationship between the semi-minor axis squared ($b^2$) and the semi-major axis ($a$).

Step 2: Using the Distance Between a Focus and its Nearest Vertex

The problem states that the distance between a focus and its nearest vertex on the major axis is $\frac{3}{2}$ units.

• Identify the coordinates:
  • The foci are at $(\pm ae, 0)$. Let's consider the positive focus $(ae, 0)$.
  • The vertices on the major axis are $(\pm a, 0)$.
• Determine the nearest vertex: Since $0 < e < 1$, we know that $ae < a$. Therefore, the vertex $(a, 0)$ is closer to the focus $(ae, 0)$ than the vertex $(-a, 0)$.
• Calculate the distance: The distance between $(ae, 0)$ and $(a, 0)$ is the absolute difference of their x-coordinates: $\text{Distance} = |a - ae|$ Since $a$ is a length, $a > 0$. Also, since $0 < e < 1$, $1 - e > 0$. Thus, $a - ae$ is positive. $\text{Distance} = a(1 - e)$
• Set up the equation: We are given this distance is $\frac{3}{2}$ units. $a(1 - e) = \frac{3}{2} \quad \text{(Equation 2)}$ This equation provides a relationship between the semi-major axis ($a$) and the eccentricity ($e$).

Step 3: Connecting the Equations using the Fundamental Relation

Now we have two equations (Equation 1 and Equation 2) involving $a$, $b$, and $e$. We need to find $e$. The key is to use the fundamental relation for an ellipse, $b^2 = a^2(1 - e^2)$, to eliminate $b$.

• Recall the fundamental relation: $b^2 = a^2(1 - e^2)$
• Substitute Equation 1 into the fundamental relation: We know $b^2 = 2a$ from Equation 1. Substitute this into the fundamental relation: $2a = a^2(1 - e^2)$
• Simplify the equation: Since $a$ is the length of the semi-major axis, $a \neq 0$. Therefore, we can safely divide both sides by $a$: $2 = a(1 - e^2)$
• Factor the term $(1 - e^2)$: Recognize that $(1 - e^2)$ is a difference of squares, which can be factored as $(1 - e)(1 + e)$. $2 = a(1 - e)(1 + e) \quad \text{(Equation 3)}$ This new equation now relates $a$ and $e$, just like Equation 2.

Step 4: Solving for the Eccentricity ($e$)

We now have a system of two equations involving only $a$ and $e$:

1. $a(1 - e) = \frac{3}{2}$ (from Step 2)
2. $a(1 - e)(1 + e) = 2$ (from Step 3)

Notice that the term $a(1 - e)$ appears in both equations. This is a common strategy in such problems: look for common expressions to simplify the system.

• Substitute Equation 2 into Equation 3: Replace $a(1 - e)$ in Equation 3 with $\frac{3}{2}$ from Equation 2. $\left(\frac{3}{2}\right)(1 + e) = 2$
• Solve for $e$: Multiply both sides by 2 to clear the fraction: $3(1 + e) = 4$ Distribute the 3: $3 + 3e = 4$ Subtract 3 from both sides: $3e = 4 - 3$ $3e = 1$ Divide by 3: $e = \frac{1}{3}$

Verification and Final Answer

The calculated eccentricity is $e = \frac{1}{3}$. This value is between 0 and 1, which is consistent with the definition of eccentricity for an ellipse.

To be thorough, we could also find $a$ and $b$. From $a(1-e) = \frac{3}{2}$: $a\left(1 - \frac{1}{3}\right) = \frac{3}{2}$ $a\left(\frac{2}{3}\right) = \frac{3}{2}$ $a = \frac{3}{2} \times \frac{3}{2} = \frac{9}{4}$

Then from $b^2 = 2a$: $b^2 = 2\left(\frac{9}{4}\right) = \frac{9}{2}$ So, $b = \frac{3}{\sqrt{2}}$. Since $a = \frac{9}{4} = 2.25$ and $b = \frac{3}{\sqrt{2}} \approx 2.12$, we have $a > b$, confirming it's a valid ellipse.

The eccentricity of the ellipse is $\frac{1}{3}$.

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Key Takeaway

This problem beautifully illustrates how to combine different geometric properties of an ellipse to solve for an unknown parameter. The strategy involved:

1. Translating given verbal information into algebraic equations using standard formulas.
2. Using the fundamental relation $b^2 = a^2(1 - e^2)$ as a bridge to connect equations involving $a$, $b$, and $e$.
3. Strategically substituting expressions to eliminate variables and simplify the system of equations.

Always remember the range of eccentricity for an ellipse ($0 < e < 1$) as a quick check for your answer!

The final answer is \boxed{\text{1 \over 3}}.