This problem masterfully combines concepts from coordinate geometry, specifically ellipses, with function optimization using quadratic functions. To tackle it effectively, we need to systematically apply the fundamental definitions and formulas for an ellipse and the techniques for finding the minimum value of a quadratic function.

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1. Understanding the Ellipse and its Core Properties

An ellipse is a conic section characterized by its distinct oval shape. For an ellipse centered at the origin, with its major axis lying along the x-axis, its standard equation is given by: $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ Here, $a$ represents the length of the semi-major axis (half the length of the major axis), and $b$ represents the length of the semi-minor axis (half the length of the minor axis). By convention, for this standard form, we assume $a > b$.

Two critical properties of an ellipse that are directly relevant to this problem are:

• Length of the Latus Rectum (L.R.): The latus rectum is a chord that passes through one of the foci and is perpendicular to the major axis. Its length provides a direct relationship between the semi-major and semi-minor axes. For an ellipse with the major axis along the x-axis ($a > b$), the length of the latus rectum is: $\text{L.R.} = \frac{2b^2}{a}$
• Eccentricity ($e$): Eccentricity is a crucial parameter that quantifies how "stretched out" or "circular" an ellipse is. For an ellipse, its value always lies strictly between 0 and 1 ($0 < e < 1$). It connects the semi-major axis ($a$) and semi-minor axis ($b$) through the formula: $e^2 = 1 - \frac{b^2}{a^2}$ This formula can also be rearranged to express $b^2$ in terms of $a$ and $e$: $b^2 = a^2(1 - e^2)$.

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2. Step 1: Establishing a Relationship from the Latus Rectum Length

The problem provides a direct piece of information: the length of the latus rectum of the given ellipse is 10. Our first step is to use the formula for the length of the latus rectum to create an equation relating $a$ and $b^2$.

Key Concept: The length of the latus rectum for an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is $\frac{2b^2}{a}$.

Why this step? The length of the latus rectum is a given numerical value, and its formula is a fundamental property of an ellipse. By substituting the given value, we can establish our first algebraic relationship between the unknown dimensions $a$ and $b^2$ of the ellipse. This is crucial for eventually solving for these dimensions.

Substitute the given value L.R. = 10 into the formula: $10 = \frac{2b^2}{a}$ Now, we simplify this equation to express $b^2$ in terms of $a$, which will be convenient for substitution later: $10a = 2b^2$ $b^2 = 5a \quad \text{(Equation 1)}$ This equation is a foundational link between $a$ and $b^2$, derived directly from the problem statement.

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3. Step 2: Calculating the Eccentricity from the Given Function

The problem states that the eccentricity $e$ of the ellipse is equal to the minimum value of the quadratic function $f(t) = t^2 + t + \frac{11}{12}$, for $t \in \mathbb{R}$.

Key Concept: The minimum value of a quadratic function $f(t) = At^2 + Bt + C$ (where $A > 0$) occurs at its vertex, $t = -\frac{B}{2A}$. The minimum value is $f\left(-\frac{B}{2A}\right)$.

Why this step? To find the eccentricity $e$, we must first determine the minimum value of the given quadratic function. This value will then be assigned to $e$.

The given function is $f(t) = t^2 + t + \frac{11}{12}$. This is a quadratic function in the form $At^2 + Bt + C$, with coefficients $A=1$, $B=1$, and $C=\frac{11}{12}$. Since the coefficient of $t^2$ ($A=1$) is positive, the parabola opens upwards, confirming that it has a global minimum.

We can find this minimum using two common methods:

1. Using the vertex formula: The $t$-coordinate of the vertex of a parabola $At^2 + Bt + C$ is $t = -\frac{B}{2A}$. Substituting $A=1$ and $B=1$: $t = -\frac{1}{2(1)} = -\frac{1}{2}$ This is the value of $t$ at which the function attains its minimum.

2. Completing the square: We can rewrite the function by completing the square for the terms involving $t$: $f(t) = (t^2 + t) + \frac{11}{12}$ To complete the square for $t^2 + t$, we add and subtract $\left(\frac{1}{2} \times \text{coefficient of } t\right)^2 = \left(\frac{1}{2} \times 1\right)^2 = \frac{1}{4}$: $f(t) = \left(t^2 + t + \frac{1}{4}\right) - \frac{1}{4} + \frac{11}{12}$ $f(t) = \left(t + \frac{1}{2}\right)^2 - \frac{3}{12} + \frac{11}{12}$ $f(t) = \left(t + \frac{1}{2}\right)^2 + \frac{8}{12}$ $f(t) = \left(t + \frac{1}{2}\right)^2 + \frac{2}{3}$ The term $\left(t + \frac{1}{2}\right)^2$ is always non-negative. Its minimum value is 0, which occurs when $t = -\frac{1}{2}$. Thus, the minimum value of $f(t)$ is $0 + \frac{2}{3} = \frac{2}{3}$.

Now, substitute $t = -\frac{1}{2}$ into the function $f(t)$ to find its minimum value: $f\left(-\frac{1}{2}\right) = \left(-\frac{1}{2}\right)^2 + \left(-\frac{1}{2}\right) + \frac{11}{12}$ $f\left(-\frac{1}{2}\right) = \frac{1}{4} - \frac{1}{2} + \frac{11}{12}$ To sum these fractions, we find a common denominator, which is 12: $f\left(-\frac{1}{2}\right) = \frac{3}{12} - \frac{6}{12} + \frac{11}{12}$ $f\left(-\frac{1}{2}\right) = \frac{3 - 6 + 11}{12} = \frac{8}{12} = \frac{2}{3}$ Therefore, the minimum value of the function $f(t)$ is $\frac{2}{3}$.

According to the problem statement, this minimum value is the eccentricity $e$ of the ellipse: $e = \frac{2}{3}$ Important Check: For an ellipse, the eccentricity $e$ must always satisfy $0 < e < 1$. Our calculated value $e = \frac{2}{3}$ perfectly fits this condition, confirming its validity. Since the eccentricity formula uses $e^2$, we calculate that value: $e^2 = \left(\frac{2}{3}\right)^2 = \frac{4}{9}$

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4. Step 3: Connecting Eccentricity to the Semi-axes $a$ and $b$

Now that we have the value of $e^2$, we can use the fundamental relationship between the eccentricity and the semi-axes of an ellipse.

Key Concept: The eccentricity $e$ of an ellipse relates its semi-major