To solve the problem step-by-step, we will first derive the equation of the ellipse from the given parametric equations, find the foci, determine the coordinates of point B (one of the ends of the latus rectum), and finally calculate the area of triangle BSS'.
### Step 1: Derive the equation of the ellipse
Given the parametric equations:
\[ x = 3 + 5 \cos \theta \]
\[ y = -2 + 4 \sin \theta \]
We can express \(\cos \theta\) and \(\sin \theta\) in terms of \(x\) and \(y\):
\[ \cos \theta = \frac{x - 3}{5} \]
\[ \sin \theta = \frac{y + 2}{4} \]
Using the identity \(\sin^2 \theta + \cos^2 \theta = 1\):
\[
\left(\frac{x - 3}{5}\right)^2 + \left(\frac{y + 2}{4}\right)^2 = 1
\]
This simplifies to:
\[
\frac{(x - 3)^2}{25} + \frac{(y + 2)^2}{16} = 1
\]
### Step 2: Identify the center, semi-major axis, and semi-minor axis
From the equation, we can identify:
- Center: \((h, k) = (3, -2)\)
- Semi-major axis \(a = 5\) (along the x-axis)
- Semi-minor axis \(b = 4\) (along the y-axis)
### Step 3: Find the foci of the ellipse
The distance \(c\) from the center to each focus is given by:
\[
c = \sqrt{a^2 - b^2} = \sqrt{25 - 16} = \sqrt{9} = 3
\]
Thus, the foci \(S\) and \(S'\) are located at:
\[
S(3 - 3, -2) = (0, -2) \quad \text{and} \quad S'(3 + 3, -2) = (6, -2)
\]
### Step 4: Find the coordinates of point B (end of the latus rectum)
The latus rectum of an ellipse is vertical at the foci. The coordinates of the ends of the latus rectum can be found using:
\[
\left(h \pm c, k \pm \frac{b^2}{a}\right)
\]
Calculating the y-coordinates:
\[
\frac{b^2}{a} = \frac{16}{5}
\]
Thus, the coordinates of point B (one of the ends of the latus rectum) are:
\[
B(0, -2 \pm \frac{16}{5}) = \left(0, -2 + \frac{16}{5}\right) \text{ or } \left(0, -2 - \frac{16}{5}\right)
\]
Calculating:
\[
-2 + \frac{16}{5} = -\frac{10}{5} + \frac{16}{5} = \frac{6}{5}
\]
\[
-2 - \frac{16}{5} = -\frac{10}{5} - \frac{16}{5} = -\frac{26}{5}
\]
So, the coordinates of point B are:
\[
B(0, \frac{6}{5}) \text{ or } B(0, -\frac{26}{5})
\]
### Step 5: Calculate the area of triangle BSS'
The area \(A\) of triangle BSS' can be calculated using the formula:
\[
A = \frac{1}{2} \times \text{base} \times \text{height}
\]
Here, the base \(SS'\) is the distance between the foci:
\[
SS' = 6 - 0 = 6
\]
The height is the vertical distance from point B to the line joining S and S':
For \(B(0, \frac{6}{5})\):
\[
\text{height} = \left| \frac{6}{5} - (-2) \right| = \left| \frac{6}{5} + \frac{10}{5} \right| = \left| \frac{16}{5} \right| = \frac{16}{5}
\]
Calculating the area:
\[
A = \frac{1}{2} \times 6 \times \frac{16}{5} = \frac{48}{5}
\]
### Final Answer
The area of triangle BSS' is:
\[
\boxed{\frac{48}{5}} \text{ square units}
\]
