Understanding the Goal and Key Formula

The problem asks us to find the length of the latus rectum of an ellipse. The latus rectum is a chord passing through a focus and perpendicular to the major axis. For any ellipse, regardless of its orientation (whether its major axis is horizontal or vertical), the length of the latus rectum, denoted by $L$, is given by the formula:

$L = \frac{2b^2}{a}$

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).

Our primary strategy will be to use the given information – the coordinates of the foci and the eccentricity – to first determine the values of $a$ and $b^2$, and then substitute these values into the latus rectum formula.

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Step 1: Determine the Center and Orientation of the Ellipse

The given foci are $F_1(2,5)$ and $F_2(2,-3)$.

• Why this step is important: The foci are fundamental to defining an ellipse. Their positions allow us to locate the center of the ellipse and determine the orientation of its major axis. This information is crucial for correctly interpreting distances and applying formulas.

• Observation about Foci: Notice that the x-coordinates of both foci are identical ($x=2$). This significant observation tells us that the major axis of the ellipse must be a vertical line, specifically the line $x=2$. If the y-coordinates were the same, the major axis would be horizontal.

• Calculating the Center: The center of an ellipse is always the midpoint of the segment connecting its two foci. Let the center of the ellipse be $(h,k)$. Using the midpoint formula for $F_1(2,5)$ and $F_2(2,-3)$: $(h,k) = \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$ $(h,k) = \left(\frac{2+2}{2}, \frac{5+(-3)}{2}\right)$ $(h,k) = \left(\frac{4}{2}, \frac{2}{2}\right)$ $(h,k) = (2,1)$ So, the center of the ellipse is $(2,1)$.

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Step 2: Calculate the Length of the Semi-Major Axis ($a$)

We are given the foci $F_1(2,5)$ and $F_2(2,-3)$, and the eccentricity $e = \frac{4}{5}$.

• Key Concept: A fundamental property of an ellipse is that the distance between its two foci is equal to $2ae$, where $a$ is the length of the semi-major axis and $e$ is the eccentricity. This relationship directly links the physical separation of the foci to the ellipse's size ($a$) and shape ($e$).

• Calculating the Distance Between Foci: Since the x-coordinates of the foci are the same, the distance between $F_1(2,5)$ and $F_2(2,-3)$ is simply the absolute difference of their y-coordinates: $\text{Distance}(F_1, F_2) = |5 - (-3)| = |5+3| = 8$ (Alternatively, using the distance formula: $\sqrt{(2-2)^2 + (5-(-3))^2} = \sqrt{0^2 + 8^2} = \sqrt{64} = 8$).

• Using the $2ae$ Relationship: Now, we equate the calculated distance between foci to $2ae$: $2ae = 8$ We are given the eccentricity $e = \frac{4}{5}$. Substitute this value into the equation: $2a\left(\frac{4}{5}\right) = 8$ $\frac{8a}{5} = 8$ To solve for $a$, we multiply both sides of the equation by $\frac{5}{8}$: $a = 8 \times \frac{5}{8}$ $a = 5$ Thus, the length of the semi-major axis is $5$.

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Step 3: Calculate the Square of the Length of the Semi-Minor Axis ($b^2$)

We have determined $a=5$ and are given $e=\frac{4}{5}$.

• Key Concept: For any ellipse, the semi-major axis ($a$), semi-minor axis ($b$), and eccentricity ($e$) are related by the fundamental equation: $b^2 = a^2(1-e^2)$ This formula is derived from the Pythagorean relationship $c^2 = a^2 - b^2$, where $c$ is the distance from the center to a focus, and $c=ae$. Substituting $c=ae$ into the Pythagorean relationship gives $a^2e^2 = a^2 - b^2$, which rearranges to $b^2 = a^2 - a^2e^2 = a^2(1-e^2)$. This equation is vital for finding $b$ (or $b^2$) when $a$ and $e$ are known.

• Calculating $b^2$: Substitute the values of $a=5$ and $e=\frac{4}{5}$ into the formula: $b^2 = (5)^2 \left(1 - \left(\frac{4}{5}\right)^2\right)$ $b^2 = 25 \left(1 - \frac{16}{25}\right)$ To simplify the term in the parenthesis, find a common denominator: $b^2 = 25 \left(\frac{25}{25} - \frac{16}{25}\right)$ $b^2 = 25 \left(\frac{25-16}{25}\right)$ $b^2 = 25 \left(\frac{9}{25}\right)$ $b^2 = 9$ We have now found $b^2 = 9$. It's important to note that we need $b^2$ for the latus rectum formula, so there's no need to calculate $b = \sqrt{9} = 3$.

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Step 4: Calculate the Length of the Latus Rectum ($L$)

We have successfully found $a=5$ and $b^2=9$.

• Key Concept: This is the final step, where we directly apply the formula for the length of the latus rectum, $L = \frac{2b^2}{a}$, using the values we've derived.

• Final Calculation: Substitute the values of $a=5$ and $b^2=9$ into the formula: $L = \frac{2 \times 9}{5}$ $L = \frac{18}{5}$

Therefore, the length of the latus rectum of the ellipse is $\frac{18}{5}$.

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Tips and Common Mistakes to Avoid:

• Identify Orientation Early: Always start by determining the center and orientation of the major axis from the foci. This prevents errors in setting up equations or interpreting coordinates. For example, if foci had same y-coordinate, it would be a horizontal ellipse.
• Memorize Key Formulas: Ensure you know the fundamental relationships:
  • Distance between foci: $2ae$
  • Relationship between $a, b, e$: $b^2 = a^2(1-e^2)$ (or $a^2e^2 = a^2 - b^2$)
  • Latus Rectum: $L = \frac{2b^2}{a}$
• Distinguish $a$ from $a^2$ and $b$ from $b^2$: Be careful when substituting values into formulas. It's a common mistake to use $a$ instead of $a^2$ or vice-versa, especially in the $b^2 = a^2(1-e^2)$ formula.
• Fraction Arithmetic: Be meticulous with calculations involving fractions, especially when squaring the eccentricity and subtracting from 1.

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Summary and Key Takeaway

To determine the length of the latus rectum of an ellipse when provided with its foci and eccentricity, follow this systematic approach:

1. Locate Center and Orientation: Use the midpoint of the foci to find the center and observe the common coordinate to determine the major axis's orientation.
2. Calculate Semi-Major Axis ($a$): Use the distance between the foci ($2ae$) and the given eccentricity ($e$) to solve for $a$.
3. Calculate Square of Semi-Minor Axis ($b^2$): Employ the fundamental relationship $b^2 = a^2(1-e^2)$ with the known $a$ and $e$ to find $b^2$.
4. Compute Latus Rectum Length ($L$): Substitute the calculated values of $a$ and $b^2$ into the formula $L = \frac{2b^2}{a}$.

This structured method ensures all given information is effectively used to arrive at the correct solution.

The final answer is $\boxed{\text{B}}$.