To find the equation of the locus of the centroid of the triangle with vertices \((a \cos k, a \sin k)\), \((b \sin k, -b \cos k)\), and \((1, 0)\), we will follow these steps:
### Step 1: Define the Centroid
The centroid \(G\) of a triangle with vertices \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\) is given by the formula:
\[
G\left(x, y\right) = \left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right)
\]
For our triangle, the vertices are:
- \( (x_1, y_1) = (a \cos k, a \sin k) \)
- \( (x_2, y_2) = (b \sin k, -b \cos k) \)
- \( (x_3, y_3) = (1, 0) \)
### Step 2: Calculate the Coordinates of the Centroid
Using the centroid formula:
\[
x = \frac{a \cos k + b \sin k + 1}{3}
\]
\[
y = \frac{a \sin k - b \cos k + 0}{3}
\]
### Step 3: Rearranging the Equations
We can rearrange the equations for \(x\) and \(y\):
\[
3x = a \cos k + b \sin k + 1 \quad \text{(1)}
\]
\[
3y = a \sin k - b \cos k \quad \text{(2)}
\]
### Step 4: Isolate \( \cos k \) and \( \sin k \)
From equation (1):
\[
a \cos k + b \sin k = 3x - 1 \quad \text{(3)}
\]
From equation (2):
\[
a \sin k - b \cos k = 3y \quad \text{(4)}
\]
### Step 5: Square and Add the Equations
Now, we will square equations (3) and (4) and add them:
\[
(a \cos k + b \sin k)^2 + (a \sin k - b \cos k)^2 = (3x - 1)^2 + (3y)^2
\]
Expanding both sides:
\[
a^2 \cos^2 k + 2ab \cos k \sin k + b^2 \sin^2 k + a^2 \sin^2 k - 2ab \sin k \cos k + b^2 \cos^2 k = (3x - 1)^2 + (3y)^2
\]
### Step 6: Combine Like Terms
Combining the terms gives:
\[
a^2 (\cos^2 k + \sin^2 k) + b^2 (\sin^2 k + \cos^2 k) = (3x - 1)^2 + (3y)^2
\]
Using the identity \(\cos^2 k + \sin^2 k = 1\):
\[
a^2 + b^2 = (3x - 1)^2 + (3y)^2
\]
### Step 7: Final Equation of the Locus
Rearranging gives us the equation of the locus:
\[
(3x - 1)^2 + (3y)^2 = a^2 + b^2
\]
### Conclusion
The equation of the locus of the centroid of the triangle is:
\[
(3x - 1)^2 + (3y)^2 = a^2 + b^2
\]
