To find the locus of the centroid of the triangle with vertices at \( (a \cos t, a \sin t) \), \( (b \sin t - b \cos t, 0) \), and \( (1, 0) \), we will follow these steps:
### Step 1: Identify the vertices of the triangle
The vertices of the triangle are:
- Vertex A: \( (a \cos t, a \sin t) \)
- Vertex B: \( (b \sin t - b \cos t, 0) \)
- Vertex C: \( (1, 0) \)
### Step 2: Use the formula for the centroid
The coordinates of the centroid \( (x, y) \) of a triangle with vertices \( (x_1, y_1) \), \( (x_2, y_2) \), and \( (x_3, y_3) \) are given by:
\[
x = \frac{x_1 + x_2 + x_3}{3}, \quad y = \frac{y_1 + y_2 + y_3}{3}
\]
### Step 3: Substitute the vertices into the centroid formula
Substituting the coordinates of the vertices into the formula:
- For the x-coordinate:
\[
x = \frac{a \cos t + (b \sin t - b \cos t) + 1}{3} = \frac{(a - b) \cos t + b \sin t + 1}{3}
\]
- For the y-coordinate:
\[
y = \frac{a \sin t + 0 + 0}{3} = \frac{a \sin t}{3}
\]
### Step 4: Rearranging the equations
From the equations derived:
1. \( 3x = (a - b) \cos t + b \sin t + 1 \)
2. \( 3y = a \sin t \)
### Step 5: Express \( \cos t \) and \( \sin t \) in terms of \( x \) and \( y \)
From the second equation, we can express \( \sin t \):
\[
\sin t = \frac{3y}{a}
\]
Substituting this into the first equation:
\[
3x - 1 = (a - b) \cos t + b \left(\frac{3y}{a}\right)
\]
Now, we need to express \( \cos t \) in terms of \( y \).
### Step 6: Use the Pythagorean identity
Using the identity \( \sin^2 t + \cos^2 t = 1 \):
\[
\cos^2 t = 1 - \left(\frac{3y}{a}\right)^2
\]
### Step 7: Substitute \( \cos t \) back into the equation
Substituting \( \cos t \) into the rearranged equation:
\[
3x - 1 = (a - b) \sqrt{1 - \left(\frac{3y}{a}\right)^2} + \frac{3by}{a}
\]
### Step 8: Square both sides to eliminate the square root
Squaring both sides will give us a relationship between \( x \) and \( y \).
### Step 9: Final equation of the locus
After simplifying, we will find that the locus of the centroid is given by:
\[
(3x - 1)^2 + 3y^2 = a^2 + b^2
\]
This represents a circle in the coordinate plane.
### Conclusion
The locus of the centroid of the triangle is:
\[
(3x - 1)^2 + 3y^2 = a^2 + b^2
\]
