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Step-by-Step Solution
Step 1: Identify given information
The ellipse has its center at the origin, with its major axis along the x-axis.
Its eccentricity is given as $\frac{3}{5}$, and the distance between its foci is 6.
Step 2: Relate distance between foci to ellipse parameters
For an ellipse with major axis along the x-axis, the foci are at $(\pm c, 0)$, where $c = ae$ and $a$ is the semi-major axis.
The distance between the foci is $2c$, so:
$2c = 6 \implies 2(ae) = 6 \implies a e = 3.$
Since $e = \frac{3}{5}$, we have:
$a \times \frac{3}{5} = 3.$
Therefore, $a = 5.$
Step 3: Compute the semi-minor axis b
The relationship between $a$, $b$, and $e$ for an ellipse is
$b^2 = a^2 \bigl(1 - e^2\bigr).$
Substituting $a = 5$ and $e = \tfrac{3}{5}$:
$b^2 = 25 \left(1 - \left(\frac{3}{5}\right)^2\right)
= 25 \left(1 - \frac{9}{25}\right)
= 25 \left(\frac{16}{25}\right)
= 16.$
Hence, $b = 4.$
Step 4: Interpret the quadrilateral inscribed in the ellipse
The vertices of the ellipse lie at $(\pm a, 0)$ and $(0, \pm b)$.
These four points form a rhombus whose diagonals have lengths $2a$ and $2b$.
Step 5: Calculate the area of the quadrilateral
The area of a rhombus is given by:
$ \frac{1}{2} \times (\text{diagonal}_1) \times (\text{diagonal}_2). $
Here, $\text{diagonal}_1 = 2a = 2 \times 5 = 10$ and $\text{diagonal}_2 = 2b = 2 \times 4 = 8.$
Therefore, the area is
$ \frac{1}{2} \times 10 \times 8 = 40. $
Final Answer
The area of the quadrilateral is 40.
