Understanding the Ellipse and its Properties

This problem asks us to find the area of a quadrilateral formed by the vertices of an ellipse. To solve this, we need to recall the fundamental properties of an ellipse, especially one centered at the origin with its major axis along the x-axis.

For such an ellipse:

• Its standard equation is given by: $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ where $a$ is the length of the semi-major axis (along the x-axis) and $b$ is the length of the semi-minor axis (along the y-axis). By definition, for this orientation, $a > b$.
• The vertices are the points where the ellipse intersects its axes. They are:
  • On the major axis (x-axis): $V_1(a, 0)$ and $V_2(-a, 0)$
  • On the minor axis (y-axis): $V_3(0, b)$ and $V_4(0, -b)$
• The foci are two special points inside the ellipse, located on the major axis at coordinates $F_1(ae, 0)$ and $F_2(-ae, 0)$, where $e$ is the eccentricity.
• The distance between the foci is $2ae$.
• The eccentricity ($e$) is a measure of how "flat" the ellipse is. It satisfies $0 < e < 1$. It is related to $a$ and $b$ by the equation: $b^2 = a^2(1 - e^2)$ This formula is crucial for finding $b$ if $a$ and $e$ are known.
• The quadrilateral inscribed in the ellipse with vertices as the vertices of the ellipse is formed by connecting $V_1, V_3, V_2, V_4$ in order. This specific quadrilateral is always a rhombus (or a square if $a=b$, which is a circle). Its diagonals are along the coordinate axes.
  • The length of the diagonal along the x-axis is $d_1 = 2a$.
  • The length of the diagonal along the y-axis is $d_2 = 2b$. The area of a rhombus is given by $\frac{1}{2} d_1 d_2$. Therefore, the area of this specific quadrilateral is: $\text{Area} = \frac{1}{2} (2a)(2b) = 2ab$

Problem Statement Analysis

We are given the following information:

• Ellipse center: Origin $(0,0)$.
• Major axis: Along the x-axis. This confirms the use of the formulas above.
• Eccentricity ($e$): $e = \frac{3}{5}$.
• Distance between foci: $6$ units.
• Goal: Find the area of the quadrilateral whose vertices are the vertices of the ellipse.

Our strategy will be to first use the given eccentricity and distance between foci to find the lengths of the semi-major axis ($a$) and semi-minor axis ($b$). Once $a$ and $b$ are known, we can directly calculate the area of the rhombus formed by the ellipse's vertices using the formula $2ab$.

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Step-by-Step Solution

Step 1: Determine the length of the semi-major axis ($a$)

• Why this step? The semi-major axis $a$ is a fundamental dimension of the ellipse. It is directly related to the distance between the foci and the eccentricity, both of which are given in the problem. Calculating $a$ is the first logical step towards defining the ellipse's dimensions.
• Key Concept: The distance between the foci of an ellipse with its major axis along the x-axis is $2ae$.
• Given Information:
  • Distance between foci $= 6$
  • Eccentricity $e = \frac{3}{5}$
• Applying the formula and Calculation: We set up the equation based on the given distance between foci: $2ae = 6$ Now, substitute the given value of $e$ into this equation: $2a \left(\frac{3}{5}\right) = 6$ Simplify the left side of the equation: $\frac{6a}{5} = 6$ To solve for $a$, we multiply both sides of the equation by $\frac{5}{6}$: $a = 6 \times \frac{5}{6}$ $a = 5$ Thus, the length of the semi-major axis is $5$ units.

Step 2: Calculate the length of the semi-minor axis ($b$)

• Why this step? The semi-minor axis $b$ is the other crucial dimension needed to fully define the ellipse and, specifically, to determine the coordinates of the vertices along the y-axis. It is also essential for calculating the area of the quadrilateral, which depends on both $a$ and $b$.
• Key Concept: The fundamental relationship connecting $a$, $b$, and $e$ for an ellipse is $b^2 = a^2(1 - e^2)$.
• Known Values:
  • $a = 5$ (calculated in Step 1)
  • $e = \frac{3}{5}$ (given)
• Applying the formula and Calculation: Substitute the values of $a$ and $e$ into the relationship: $b^2 = (5)^2 \left(1 - \left(\frac{3}{5}\right)^2\right)$ First, square the values: $b^2 = 25 \left(1 - \frac{9}{25}\right)$ Perform the subtraction inside the parenthesis by finding a common denominator: $b^2 = 25 \left(\frac{25 - 9}{25}\right)$ $b^2 = 25 \left(\frac{16}{25}\right)$ The $25$ in the numerator and denominator cancel out: $b^2 = 16$ To find $b$, take the square root of both sides. Since $b$ represents a length, we consider only the positive root: $b = \sqrt{16}$ $b = 4$ Therefore, the length of the semi-minor axis is $4$ units.

Step 3: Identify the Vertices of the Ellipse

• Why this step? The problem explicitly asks for the area of a quadrilateral whose vertices are the vertices of the ellipse. Listing these coordinates allows us to visualize the quadrilateral and confirm its properties (like being a rhombus with diagonals along the axes).
• Key Concept: For an ellipse centered at the origin with its major axis along the x-axis, the vertices are $(\pm a, 0)$ and $(0, \pm b)$.
• Known Values:
  • $a = 5$
  • $b = 4$
• Calculation: Using our calculated values, the four vertices of the ellipse are:
  • $V_1 = (a, 0) = (5, 0)$
  • $V_2 = (-a, 0) = (-5, 0)$
  • $V_3 = (0, b) = (0, 4)$
  • $V_4 = (0, -b) = (0, -4)$

Step 4: Calculate the Area of the Inscribed Quadrilateral

• Why this step? This is the final step to answer the question. We have all the necessary information ($a$ and $b$) to apply the area formula derived earlier.

• Key Concept: The quadrilateral formed by the vertices of the ellipse is a rhombus. Its diagonals lie along the x and y axes.

  • Length of diagonal along x-axis ($d_1$) = distance between $(5,0)$ and $(-5,0)$ = $5 - (-5) = 10$ units. This is $2a$.
  • Length of diagonal along y-axis ($d_2$) = distance between $(0,4)$ and $(0,-4)$ = $4 - (-4) = 8$ units. This is $2b$. The area of a rhombus is given by $\frac{1}{2} d_1 d_2$. Alternatively, the area of this specific quadrilateral is $2ab$.

• Known Values:

  • $a = 5$
  • $b = 4$

• Applying the formula and Calculation: Using the area formula $2ab$: $\text{Area} = 2 \times a \times b$ $\text{Area} = 2 \times 5 \times 4$ $\text{Area} = 2 \times 20$ $\text{Area} = 40 \text{ sq. units}$

  (Self-check using diagonal formula) $\text{Area} = \frac{1}{2} d_1 d_2 = \frac{1}{2} (10)(8) = \frac{1}{2} (80) = 40 \text{ sq. units}$ Both methods yield the same result, confirming our calculation.

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Important Tips and Common Pitfalls

• Understand the Orientation: Always carefully note whether the major axis is along the x-axis or y-axis. This determines the standard equation, the coordinates of vertices and foci, and which variable ($a$ or $b$) corresponds to the semi-major axis. If the major axis were along the y-axis, the equation would be $\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$, and vertices would be $(0, \pm a)$ and $(\pm b, 0)$.
• 'a' is Always Semi-Major Axis: By convention, $a$ is always defined as the length of the semi-major axis and $b$ as the length of the semi-minor axis. Therefore, $a > b$ is always true for an ellipse.
• Visualize the Quadrilateral: The quadrilateral formed by the four vertices of an ellipse (when its axes align with the coordinate axes) is always a rhombus. Its diagonals are $2a$ and $2b$. Remembering this allows for a very quick calculation of the area using $\frac{1}{2} d_1 d_2$ or simply $2ab$.
• Double-Check Calculations: Especially with squaring terms and fractions, a small arithmetic error can lead to a completely different answer. Take your time with each calculation step.

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Summary and Key Takeaway

This problem is a straightforward application of the fundamental properties of an ellipse. We systematically used the given information:

1. The eccentricity $e = \frac{3}{5}$ and the distance between foci $2ae = 6$ were combined to determine the semi-major axis $a=5$.
2. The fundamental relationship $b^2 = a^2(1 - e^2)$ was then used with the calculated $a$ and given $e$ to find the semi-minor axis $b=4$.
3. Finally, recognizing that the quadrilateral formed by the vertices of the ellipse is a rhombus with diagonals $2a$ and $2b$, its area was calculated using the formula $2ab = 2(5)(4) = 40$ sq. units.

The key takeaway is the importance of knowing the standard formulas and relationships for an ellipse, and being able to extract the necessary parameters from the problem statement to apply them correctly.

The final answer is $\boxed{\text{40}}$.