Locus of the centroid of a triangle whose vertices are `(a cos t, a sint), (bsin t,-b cos t)` and (1,0) where t is parameter is :

Options

A. `(3x-1)^(2)+(3y)^(2)=a^(2)-b^(2)`
B. `(3x-1)^(2)+(3y)^(2)=a^(2)+b^(2)`
C. `(3x+1)^(2)+(3y)^(2)=a^(2)+b^(2)`
D. `(3x+1)^(2)+(3y)^(2)=a^(2)-b^(2)`

Answer

To find the locus of the centroid of the triangle with vertices \((a \cos t, a \sin t)\), \((b \sin t, -b \cos t)\), and \((1, 0)\), we will follow these steps:

### Step 1: Identify the coordinates of the vertices
The vertices of the triangle are given as:
- \(A = (a \cos t, a \sin t)\)
- \(B = (b \sin t, -b \cos t)\)
- \(C = (1, 0)\)

### Step 2: Use the formula for 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:
\[ G\left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right) \]
Substituting the coordinates of the vertices:
\[ G\left(\frac{a \cos t + b \sin t + 1}{3}, \frac{a \sin t - b \cos t + 0}{3}\right) \]

### Step 3: Simplify the coordinates of the centroid
Let \(h\) and \(k\) represent the coordinates of the centroid:
\[ h = \frac{a \cos t + b \sin t + 1}{3} \]
\[ k = \frac{a \sin t - b \cos t}{3} \]

### Step 4: Rearrange the equations for \(h\) and \(k\)
From the equations for \(h\) and \(k\), we can express them in terms of \(t\):
1. \(3h - 1 = a \cos t + b \sin t\)
2. \(3k = a \sin t - b \cos t\)

### Step 5: Square both equations
Now, we will square both equations:
1. \((3h - 1)^2 = (a \cos t + b \sin t)^2\)
2. \((3k)^2 = (a \sin t - b \cos t)^2\)

### Step 6: Expand both squared equations
Expanding the first equation:
\[ (3h - 1)^2 = a^2 \cos^2 t + b^2 \sin^2 t + 2ab \sin t \cos t \]
Expanding the second equation:
\[ (3k)^2 = a^2 \sin^2 t + b^2 \cos^2 t - 2ab \sin t \cos t \]

### Step 7: Add the two equations
Now, we will add both expanded equations:
\[ (3h - 1)^2 + (3k)^2 = (a^2 \cos^2 t + b^2 \sin^2 t + 2ab \sin t \cos t) + (a^2 \sin^2 t + b^2 \cos^2 t - 2ab \sin t \cos t) \]
This simplifies to:
\[ (3h - 1)^2 + (3k)^2 = a^2 (\cos^2 t + \sin^2 t) + b^2 (\sin^2 t + \cos^2 t) \]
Using the identity \(\cos^2 t + \sin^2 t = 1\):
\[ (3h - 1)^2 + (3k)^2 = a^2 + b^2 \]

### Step 8: Final equation of the locus
Rearranging gives us the equation of the locus:
\[ (3h - 1)^2 + (3k)^2 = a^2 + b^2 \]
Replacing \(h\) with \(x\) and \(k\) with \(y\):
\[ (3x - 1)^2 + (3y)^2 = a^2 + b^2 \]

### Conclusion
The locus of the centroid of the triangle is given by:
\[ (3x - 1)^2 + (3y)^2 = a^2 + b^2 \]