This problem requires a thorough understanding of the fundamental properties of an ellipse, including its defining equations, geometric relationships, and the significance of its parameters. Our objective is to leverage the given information about the latus rectum and the angle subtended by the minor axis at the foci to determine the lengths of the major and minor axes, and then compute a specific expression involving these lengths.

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Key Concepts and Formulas for Ellipses

For an ellipse with its center at $(h, k)$ and its major axis along the x-axis:

1. Standard Equation: The equation is given by $\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$.
   • Here, $a$ represents the length of the semi-major axis ($a>0$).
   • $b$ represents the length of the semi-minor axis ($b>0$).
   • Crucially, for the major axis along the x-axis, we have $a > b$.
2. Length of Major Axis: $2a$
3. Length of Minor Axis: $2b$
4. Length of Latus Rectum ($L$): This is the length of a chord passing through a focus and perpendicular to the major axis. It is given by $L = \frac{2b^2}{a}$.
5. Eccentricity ($e$): A dimensionless parameter that describes how "stretched" or "circular" the ellipse is. It's related to $a$ and $b$ by the equation $b^2 = a^2(1 - e^2)$. This can be rearranged to $a^2e^2 = a^2 - b^2$, which is often more convenient in calculations involving $e$.
6. Foci: The foci are located at $(h \pm ae, k)$. The distance from the center $C(h,k)$ to each focus $F$ is $ae$.
7. Geometric Property: Angle Subtended by Minor Axis at a Focus:
   • Let the center be $C(h,k)$, an endpoint of the minor axis be $B_1(h, k+b)$, and a focus be $F_1(h+ae, k)$.
   • The minor axis connects $B_1(h, k+b)$ and $B_2(h, k-b)$.
   • The triangle $\triangle B_1 F_1 B_2$ is isosceles with $F_1 B_1 = F_1 B_2$.
   • The line segment $CF_1$ is perpendicular to the minor axis $B_1B_2$.
   • Thus, in the right-angled triangle $\triangle B_1 C F_1$:
     • The length of the side $CB_1$ is $b$ (semi-minor axis).
     • The length of the side $CF_1$ is $ae$ (distance from center to focus).
     • If the minor axis $B_1B_2$ subtends an angle $2\theta$ at $F_1$, then $\angle B_1 F_1 C = \theta$.
     • From this right triangle, we can write $\tan \theta = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{CB_1}{CF_1} = \frac{b}{ae}$. This relationship is key for problems involving this geometric condition.

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Step-by-Step Solution

Step 1: Formulate an Equation from the Latus Rectum Length

• Understanding the given information: We are given an ellipse with its center at $(1,0)$ and its major axis along the x-axis. This confirms that the standard form $\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$ applies, with $h=1$ and $k=0$. We are also given that the length of the latus rectum is $\frac{1}{2}$.

• Applying the Latus Rectum formula: The formula for the length of the latus rectum ($L$) is $L = \frac{2b^2}{a}$. We substitute the given value $L = \frac{1}{2}$: $\frac{2b^2}{a} = \frac{1}{2}$ Rearranging this equation to establish a relationship between $a$ and $b$: $4b^2 = a \quad \text{(Equation 1)}$ Why this step is taken: The latus rectum length is a direct geometric property that provides an algebraic link between the semi-major axis ($a$) and the semi-minor axis ($b$). This is our first independent equation, crucial for solving for two unknowns.

Step 2: Formulate an Equation from the Angle Subtended by the Minor Axis at the Foci

• Interpreting the angle condition: The problem states that "its minor axis subtends an angle $60^{\circ}$ at the foci." Let's consider one focus, $F_1$, and the endpoints of the minor axis, $B_1$ and $B_2$. The angle $\angle B_1 F_1 B_2 = 60^{\circ}$.

• Using the geometric property: As explained in the key concepts, the line segment connecting the center $C$ to the focus $F_1$ bisects the angle $\angle B_1 F_1 B_2$. Therefore, in the right-angled triangle $\triangle B_1 C F_1$:

  • The angle $\angle B_1 F_1 C = \frac{60^{\circ}}{2} = 30^{\circ}$.
  • The side $CB_1$ (opposite to $\angle B_1 F_1 C$) has length $b$ (semi-minor axis).
  • The side $CF_1$ (adjacent to $\angle B_1 F_1 C$) has length $ae$ (distance from center to focus, where $e$ is the eccentricity).

• Applying trigonometry: We use the tangent function in $\triangle B_1 C F_1$: $\tan(\angle B_1 F_1 C) = \frac{CB_1}{CF_1}$ $\tan 30^{\circ} = \frac{b}{ae}$ We know that $\tan 30^{\circ} = \frac{1}{\sqrt{3}}$. $\frac{1}{\sqrt{3}} = \frac{b}{ae}$

• Eliminating eccentricity ($e$): Our goal is to find $a$ and $b$, so we need to eliminate $e$. We can square both sides of the equation and then use the fundamental relationship $a^2e^2 = a^2 - b^2$. Squaring both sides: $\left(\frac{1}{\sqrt{3}}\right)^2 = \left(\frac{b}{ae}\right)^2$ $\frac{1}{3} = \frac{b^2}{a^2e^2}$ Now, substitute $a^2e^2 = a^2 - b^2$: $\frac{1}{3} = \frac{b^2}{a^2 - b^2}$ Cross-multiplying gives us our second relation between $a$ and $b$: $a^2 - b^2 = 3b^2$ $a^2 = 4b^2 \quad \text{(Equation 2)}$ Why this step is taken: This geometric property is the second independent piece of information provided. It introduces eccentricity ($e$), which needs to be eliminated using the fundamental relation $b^2 = a^2(1-e^2)$ to obtain a second equation solely in terms of $a$ and $b$.

Step 3: Solve the System of Equations for $a$ and $b$}

• Our system of equations:

  1. $a = 4b^2$
  2. $a^2 = 4b^2$

• Solving by substitution: Notice that the right-hand side of both equations is $4b^2$. This means we can equate the left-hand sides: From Equation 1, we have $a = 4b^2$. Substitute this into Equation 2: $a^2 = a$ Rearrange and solve for $a$: $a^2 - a = 0$ $a(a - 1) = 0$ Since $a$ represents the length of the semi-major axis, it must be a positive value ($a > 0$). Therefore, $a \neq 0$. This implies: $a - 1 = 0 \implies a = 1$

• Solving for $b$: Now substitute the value of $a=1$ back into Equation 1: $4b^2 = 1$ $b^2 = \frac{1}{4}$ Since $b$ represents a length, it must be positive ($b > 0$): $b = \sqrt{\frac{1}{4}} = \frac{1}{2}$ So, the semi-major axis length is $a=1$ and the semi-minor axis length is $b=\frac{1}{2}$. Why this step is taken: We now have two distinct equations relating $a$ and $b$. Solving this system is essential to find the specific values of the semi-axes, which define the dimensions of the ellipse.

Step 4: Calculate the Final Required Value

• Understanding the question: The problem asks for "the square of the sum of the lengths of its minor and major axes."

  • Length of major axis = $2a$
  • Length of minor axis = $2b$
  • Sum of lengths = $2a + 2b$
  • Required value = $(2a + 2b)^2$

• Substituting the values of $a$ and $b$: Length of major axis = $2(1) = 2$ Length of minor axis = $2\left(\frac{1}{2}\right) = 1$

  Sum of lengths = $2 + 1 = 3$

  Square of the sum of lengths = $(3)^2 = 9$

  Wait! Let's recheck the problem statement and the correct answer. The correct answer is given as 2. My calculation yields 9. Let me re-examine the equations.

  Equation 1: $a = 4b^2$ Equation 2: $a^2 = 4b^2$

  If $a = 4b^2$ and $a^2 = 4b^2$, then $a^2 = a$. $a(a-1)=0$. Since $a \neq 0$, $a=1$. If $a=1$, then $4b^2 = 1 \implies b^2 = 1/4 \implies b=1/2$. These values seem correct based on the derivation.

  Let me check the question and the "Current Solution" provided. The current solution has exactly the same steps and derivation. Step 3: Solve the System of Equations for $a$ and $b$

  1. $a = 4b^2$
  2. $a^2 = 4b^2$ ... $a=1$, $b=1/2$. Step 4: Calculate the Final Required Value $(2a + 2b)^2 = (2(1) + 2(1/2))^2 = (2+1)^2 = 3^2 = 9$.

  There might be a mismatch between the provided "Correct Answer: 2" and the problem statement/solution. Based on the problem statement and the derived $a$ and $b$ values, the answer should be 9.

  Let's re-read the problem carefully: "the square of the sum of the lengths of its minor and major axes is equal to ____________." This means $(2a + 2b)^2$. If $a=1$ and $b=1/2$, then $2a=2$ and $2b=1$. Sum is $2+1=3$. Square of sum is $3^2=9$.

  Perhaps there is a subtle misinterpretation of the latus rectum or the angle. Latus Rectum $L = 2b^2/a$. Given $L=1/2$. So $2b^2/a = 1/2 \implies 4b^2 = a$. This is correct. Minor axis subtends $60^\circ$ at the foci. $\tan(60/2) = \tan(30^\circ) = b/(ae)$. $1/\sqrt{3} = b/(ae) \implies ae = b\sqrt{3} \implies a^2e^2 = 3b^2$. Using $a^2e^2 = a^2-b^2$: $a^2-b^2 = 3b^2 \implies a^2 = 4b^2$. This is also correct.

  The derivation for $a=1, b=1/2$ is robust. The calculation $(2a+2b)^2 = (2(1) + 2(1/2))^2 = (2+1)^2 = 3^2 = 9$ is also robust.

  It is highly probable that the "Correct Answer: 2" provided in the prompt is incorrect, or the question implies something else that is not standard. Assuming the question and standard formulas are correct, the answer is 9. However, as an expert teacher, I should point out this discrepancy.

  For the purpose of *rewriting the