This solution will walk you through the problem step-by-step, explaining each concept and calculation in detail.

---

1. Understanding the Fundamental Properties of an Ellipse

An ellipse is a conic section characterized by its unique shape. For an ellipse centered at the origin with its major axis along the x-axis (as indicated by $a > b$), its standard equation is: $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ Here, $a$ is the length of the semi-major axis (half the length of the major axis), and $b$ is the length of the semi-minor axis (half the length of the minor axis).

We will use two crucial properties of the ellipse:

• Length of the Latus Rectum (LR): The latus rectum is a chord that passes through a focus and is perpendicular to the major axis. Its length provides a measure of the "width" of the ellipse at its foci. For an ellipse with the major axis along the x-axis ($a>b$), the length of the latus rectum is given by: $LR = \frac{2b^2}{a}$ Why this formula? This formula is derived from the definition of an ellipse and its coordinates. The points on the latus rectum have an x-coordinate equal to the eccentricity times the semi-major axis ($ae$), and by substituting this into the ellipse equation, the y-coordinate (half the latus rectum length) can be found.

• Eccentricity ($e$): Eccentricity is a dimensionless quantity that quantifies how "stretched out" or "circular" an ellipse is. It's defined as the ratio of the distance from the center to a focus ($c$) to the length of the semi-major axis ($a$), i.e., $e = c/a$. The relationship between $a$, $b$, and $e$ is given by: $e^2 = 1 - \frac{b^2}{a^2}$ Why this formula? This formula arises from the Pythagorean relationship $a^2 = b^2 + c^2$ in an ellipse (where $c$ is the distance from the center to a focus). Substituting $c = ae$ into this relation gives $a^2 = b^2 + (ae)^2$, which simplifies to $e^2 = 1 - b^2/a^2$. Important Note: For an ellipse, the eccentricity $e$ must always satisfy $0 < e < 1$. This condition is a vital check for our calculated value of $e$.

---

2. Determining the Eccentricity ($e$) of the Ellipse

The problem states that the eccentricity ($e$) of the ellipse is equal to the maximum value of the function $\phi \left( t \right) = {5 \over {12}} + t - {t^2}$. This is a quadratic function of $t$.

Step 1: Identify the type of quadratic function. The given function is $\phi \left( t \right) = -{t^2} + t + {5 \over {12}}$. This is a quadratic function in the form $At^2 + Bt + C$, where $A = -1$, $B = 1$, and $C = 5/12$. Why is this important? Since the coefficient of $t^2$ (i.e., $A$) is negative ($-1 < 0$), the parabola represented by this function opens downwards, meaning it has a unique maximum value.

Step 2: Find the maximum value of the quadratic function. We can find the maximum value by completing the square or by using the vertex formula. Let's demonstrate by completing the square, as it's a fundamental algebraic technique.

To complete the square for $\phi(t) = -t^2 + t + 5/12$: First, factor out the coefficient of $t^2$ from the terms involving $t$: $\phi \left( t \right) = - \left( {{t^2} - t - {5 \over {12}}} \right)$ Now, focus on the quadratic expression inside the parenthesis, $t^2 - t$. To complete the square for an expression of the form $x^2 + Bx$, we add $(B/2)^2$. Here, $B = -1$, so $(B/2)^2 = (-1/2)^2 = 1/4$. We add and subtract this value inside the parenthesis: $\phi \left( t \right) = - \left( {\left( {{t^2} - t + {1 \over 4}} \right) - {1 \over 4} - {5 \over {12}}} \right)$ The terms in the first parenthesis now form a perfect square: $t^2 - t + 1/4 = (t - 1/2)^2$. $\phi \left( t \right) = - \left( {{{\left( {t - {1 \over 2}} \right)}^2} - {1 \over 4} - {5 \over {12}}} \right)$ Combine the constant terms within the parenthesis. To do this, find a common denominator (12): ${1 \over 4} = {3 \over {12}}$ So, $\phi \left( t \right) = - \left( {{{\left( {t - {1 \over 2}} \right)}^2} - {3 \over {12}} - {5 \over {12}}} \right)$ $\phi \left( t \right) = - \left( {{{\left( {t - {1 \over 2}} \right)}^2} - {8 \over {12}}} \right)$ Finally, distribute the negative sign back into the parenthesis: $\phi \left( t \right) = {8 \over {12}} - {\left( {t - {1 \over 2}} \right)^2}$ Simplify the constant term: ${8 \over {12}} = {2 \over 3}$. $\phi \left( t \right) = {2 \over 3} - {\left( {t - {1 \over 2}} \right)^2}$

Why does this give the maximum? The term ${\left( {t - {1 \over 2}} \right)^2}$ is a square, so it is always greater than or equal to zero, i.e., ${\left( {t - {1 \over 2}} \right)^2} \ge 0$. Therefore, $-{\left( {t - {1 \over 2}} \right)^2}$ is always less than or equal to zero, i.e., $-{\left( {t - {1 \over 2}} \right)^2} \le 0$. The maximum value of $\phi(t)$ occurs when $-{\left( {t - {1 \over 2}} \right)^2}$ is at its maximum, which is $0$. This happens when $t - 1/2 = 0$, or $t = 1/2$. Thus, the maximum value of $\phi(t)$ is: $\phi {\left( t \right)_{\max }} = {2 \over 3} - 0 = {2 \over 3}$

Alternative Method (Vertex Formula): For a quadratic $At^2 + Bt + C$, the maximum (or minimum) occurs at $t = -B/(2A)$. Here, $A = -1$, $B = 1$. So, $t = -1/(2 \times -1) = -1/(-2) = 1/2$. Substitute $t = 1/2$ into $\phi(t)$: $\phi \left( {1 \over 2} \right) = {5 \over {12}} + {1 \over 2} - {\left( {{1 \over 2}} \right)^2} = {5 \over {12}} + {1 \over 2} - {1 \over 4}$ Find a common denominator (12): $\phi \left( {1 \over 2} \right) = {5 \over {12}} + {6 \over {12}} - {3 \over {12}} = {{5 + 6 - 3} \over {12}} = {8 \over {12}} = {2 \over 3}$ Both methods yield the same maximum value.

Step 3: Assign the maximum value to eccentricity. The problem states that the eccentricity $e$ is this maximum value: $e = {2 \over 3}$ Check: Our calculated eccentricity $e = 2/3$ satisfies $0 < e < 1$, which is consistent for an ellipse. This confirms our calculation is valid for an ellipse.

---

3. Establishing a Relationship between $a$ and $b$ using Eccentricity

Now that we have the value of the eccentricity, we can use the fundamental eccentricity formula to establish a relationship between $a$ and $b$.

Step 4: Substitute the value of $e$ into the eccentricity formula. The formula is ${e^2} = 1 - {{{b^2}} \over {{a^2}}}$. Substitute $e = 2/3$: ${\left( {{2 \over 3}} \right)^2} = 1 - {{{b^2}} \over {{a^2}}}$ ${4 \over 9} = 1 - {{{b^2}} \over {{a^2}}}$

Step 5: Solve for the ratio ${{{b^2}} \over {{a^2}}}$. Rearrange the equation to isolate ${{{b^2}} \over {{a^2}}}$: ${{{b^2}} \over {{a^2}}} = 1 - {4 \over 9}$ ${{{b^2}} \over {{a^2}}} = {9 \over 9} - {4 \over 9}$ ${{{b^2}} \over {{a^2}}} = {5 \over 9} \quad \text{(Equation 1)}$ Why this step? This equation provides a critical relationship between the squares of the semi-major and semi-minor axes. We cannot solve for $a$ or $b$ individually yet, but we have a ratio that will be useful when combined with other information.

---

4. Utilizing the Latus Rectum Information

The problem also gives us the length of the latus rectum, which provides another equation relating $a$ and $b$.

Step 6: Use the given length of the latus rectum. The length of the latus rectum is given as $10$. Using the formula ${{2{b^2}} \over a}$: ${{2{b^2}} \over a} = 10$ Why this step? We need to find individual values for $a$ and $b^2$. Since Equation (1) provides one relationship, we need a second independent equation to form a system of equations that can be solved.

Simplify this equation: ${{{b^2}} \over a} = {10 \over 2}$ ${{{b^2}} \over a} = 5 \quad \text{(Equation 2)}$ This equation gives us another relationship between $a$ and $b^2$.

---

5. Solving for $a$ and $b^2$

We now have a system of two equations (Equation 1 and Equation 2) with two unknowns ($a$ and $b^2$). We can solve this system to find $a$ and $b^2$.

Step 7: Combine Equation 1 and Equation 2 to find 'a'. Equation 1 is: ${{{b^2}} \over {{a^2}}} = {5 \over 9}$ We can rewrite the left side of Equation 1 to incorporate the term present in Equation 2 ($b^2/a$): ${{{b^2}} \over a} \times {1 \over a} = {5 \over 9}$ Now, substitute the value of ${{{b^2}} \over a}$ from Equation 2 (which is $5$) into this modified Equation