This solution will guide you through the process of solving the problem by first understanding the underlying concepts of an ellipse, then methodically applying them step-by-step.

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Key Concepts and Formulas for an Ellipse Centered at the Origin

An ellipse is a locus of points such that the sum of the distances from two fixed points (foci) is constant. For an ellipse centered at the origin, its properties are determined by the orientation of its major axis.

Let $a_{major}$ be the length of the semi-major axis (half the length of the major axis) and $a_{minor}$ be the length of the semi-minor axis (half the length of the minor axis). The distance from the center to each focus is denoted by $c$. These three quantities are related by the fundamental equation: $c^2 = a_{major}^2 - a_{minor}^2$ The eccentricity, $e$, is defined as $e = c / a_{major}$. It measures how "squashed" the ellipse is.

There are two primary orientations for an ellipse centered at the origin:

1. Major Axis along the X-axis:

   • Equation: $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, where $a > b$.
     • Here, $a_{major} = a$ (the semi-axis associated with $x^2$), and $a_{minor} = b$ (the semi-axis associated with $y^2$).
   • Length of Major Axis: $2a$
   • Length of Minor Axis: $2b$
   • Coordinates of Foci: $(\pm c, 0) = (\pm ae, 0)$
   • Eccentricity Relation: $c = ae$, so $(ae)^2 = a^2 - b^2$, or $e^2 = 1 - \frac{b^2}{a^2}$.
   • Length of Latus Rectum: $\frac{2b^2}{a}$

2. Major Axis along the Y-axis:

   • Equation: $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, where $b > a$.
     • Here, $a_{major} = b$ (the semi-axis associated with $y^2$), and $a_{minor} = a$ (the semi-axis associated with $x^2$).
   • Length of Major Axis: $2b$
   • Length of Minor Axis: $2a$
   • Coordinates of Foci: $(0, \pm c) = (0, \pm be)$
   • Eccentricity Relation: $c = be$, so $(be)^2 = b^2 - a^2$, or $e^2 = 1 - \frac{a^2}{b^2}$.
   • Length of Latus Rectum: $\frac{2a^2}{b}$

The initial and most crucial step in solving any ellipse problem is to correctly identify its orientation, as this dictates which set of specific formulas to apply.

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Step-by-Step Solution

1. Determine the Orientation of the Ellipse and Identify Relevant Formulas.

• Given Information: We are given that one of the foci of the ellipse is at $(0, 5\sqrt{3})$.
• Why this is important: The coordinates of the foci are the key to determining the orientation of the major axis. Foci always lie on the major axis.
  • If the foci have non-zero x-coordinates and zero y-coordinates (e.g., $(\pm c, 0)$), the major axis is along the x-axis.
  • If the foci have zero x-coordinates and non-zero y-coordinates (e.g., $(0, \pm c)$), the major axis is along the y-axis.
• Conclusion: Since the x-coordinate of the given focus $(0, 5\sqrt{3})$ is zero, the foci lie on the Y-axis. Therefore, the major axis of the ellipse must lie along the Y-axis.
• Implication for semi-axes: For an ellipse with its major axis along the Y-axis, the semi-major axis is denoted by $b$ (the denominator under $y^2$ in the standard equation) and the semi-minor axis is denoted by $a$ (the denominator under $x^2$). This implies that $b > a$.
• Relevant Formulas for this Ellipse Type (Major Axis along Y-axis):
  • Foci: $(0, \pm be)$
  • Length of Major Axis: $2b$
  • Length of Minor Axis: $2a$
  • Eccentricity relation: $b^2 e^2 = b^2 - a^2$ (This comes from $c^2 = a_{major}^2 - a_{minor}^2$, where $c=be$, $a_{major}=b$, $a_{minor}=a$)
  • Length of Latus Rectum: $\frac{2a^2}{b}$

2. Utilize the Focus Information to Formulate an Equation.

• Given: One focus is at $(0, 5\sqrt{3})$.
• From Step 1: For an ellipse with its major axis along the Y-axis, the foci are $(0, \pm be)$.
• Equating: By comparing the given focus with the general form, we can establish the value of $be$: $be = 5\sqrt{3}$
• Why square this equation? Our goal is to find $a$ and $b$. The eccentricity $e$ is an auxiliary parameter that we typically want to eliminate. We know the relationship $b^2 e^2 = b^2 - a^2$. Squaring the equation for $be$ allows us to directly substitute and eliminate $e$, giving us an equation solely in terms of $a$ and $b$. $(be)^2 = (5\sqrt{3})^2$ $b^2 e^2 = 25 \times 3$ $b^2 e^2 = 75$
• Substitute Eccentricity Relation: Now, substitute the identity $b^2 e^2 = b^2 - a^2$ into the equation: $b^2 - a^2 = 75 \quad \text{(Equation 1)}$ This equation provides a relationship between the semi-major and semi-minor axes.

3. Utilize the Difference of Axis Lengths Information to Formulate Another Equation.

• Given Information: The difference of the lengths of the major axis and minor axis is 10.
• From Step 1: For this ellipse (major axis along Y-axis), the length of the major axis is $2b$ and the length of the minor axis is $2a$.
• Why this order of subtraction ($2b - 2a$)? Since $b$ is the semi-major axis, $2b$ is the length of the major axis, which is always longer than the minor axis $2a$. Therefore, the difference must be positive, so we write $2b - 2a$.
• Formulating the Equation: $2b - 2a = 10$
• Simplify: Divide the entire equation by 2 to obtain a simpler linear relationship between $a$ and $b$: $b - a = 5 \quad \text{(Equation 2)}$

4. Solve the System of Equations for Semi-Axes $a$ and $b$.

We now have a system of two equations with two unknowns ($a$ and $b$):

1. $b^2 - a^2 = 75$ (Equation 1)
2. $b - a = 5$ (Equation 2)

• Strategy: Equation 1 is in the form of a difference of squares, which can be factored as $(b-a)(b+a)$. This is a very common and effective algebraic technique to simplify such systems when one equation involves squares and the other involves a linear difference or sum.
• Factor Equation 1: $(b-a)(b+a) = 75$
• Substitute from Equation 2: We know from Equation 2 that $b-a = 5$. Substitute this value into the factored equation: $5(b+a) = 75$
• Solve for $(b+a)$: Divide both sides by 5: $b+a = \frac{75}{5}$ $b+a = 15 \quad \text{(Equation 3)}$
• Solve the system of linear equations (Equation 2 and Equation 3): Now we have a simple system of two linear equations:
  • $b - a = 5 \quad \text{(Equation 2)}$
  • $b + a = 15 \quad \text{(Equation 3)}$
  • Add Equation 2 and Equation 3: This operation will eliminate $a$: $(b - a) + (b + a) = 5 + 15$ $2b = 20$ $b = 10$
  • Substitute $b=10$ into Equation 2: $10 - a = 5$ $a = 10 - 5$ $a = 5$
• Verification: We found $b=10$ and $a=5$. This satisfies the condition $b > a$ (i.e., $10 > 5$), which is consistent with our initial determination that the major axis is along the Y-axis. If we had found $a > b$, it would indicate an error in our initial orientation assumption or calculations.

5. Calculate the Length of the Latus Rectum.

• Goal: The final step is to calculate the length of the latus rectum.
• From Step 1: For an ellipse with its major axis along the Y-axis, the length of the latus rectum is given by the formula $\frac{2a^2}{b}$.
• Substitute the values of $a$ and $b$: We found $a=5$ and $b=10$. $\text{Length of Latus Rectum} = \frac{2(5)^2}{10}$ $= \frac{2 \times 25}{10}$ $= \frac{50}{10}$ $= 5$

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Final Answer:

The length of the latus rectum is 5.

The correct option is (A).

Key Takeaways and Tips:

• Orientation First: Always start by correctly identifying the orientation of the major axis (X-axis or Y-axis) based on the given information (usually foci or vertices). This is the most critical step as it determines which set of formulas to use for $a, b$, foci, and latus rectum.
• Standard Notation vs. Semi-axes: Be careful with the use of $a$ and $b$. In the standard equation $\frac{x^2}{A^2} + \frac{y^2}{B^2} = 1$, $A$ and $B$ are the semi-axes. The larger of $A$ and $B$ is always the semi-major axis, and the smaller is the semi-minor axis. If the major axis is along the x-axis, $A$ is the semi-major axis (often denoted $a$), and $B$ is the semi-minor axis (often denoted $b$). If the major axis is along the y-axis, $B$ is the semi-major axis (often denoted $b$), and $A$ is the semi-minor axis (often denoted $a$). In this solution, we followed the convention where $b$ is the semi-major axis and $a$ is the semi-minor axis because the major axis is along the y-axis.
• Algebraic Identities: The difference of squares identity, $X^2 - Y^2 = (X-Y