The locus of the represented by` x = t^ 2 + t + 1 , y = t ^ 2 - t + 1 ` is

Answer

To find the locus represented by the equations \( x = t^2 + t + 1 \) and \( y = t^2 - t + 1 \), we will eliminate the parameter \( t \) step by step.

### Step 1: Rewrite the equations
We have:
\[ x = t^2 + t + 1 \quad (1) \]
\[ y = t^2 - t + 1 \quad (2) \]

### Step 2: Add the two equations
Let's add equations (1) and (2):
\[ x + y = (t^2 + t + 1) + (t^2 - t + 1) \]
This simplifies to:
\[ x + y = 2t^2 + 2 \]
Dividing the entire equation by 2 gives:
\[ \frac{x + y}{2} = t^2 + 1 \quad (3) \]

### Step 3: Subtract the two equations
Now, let's subtract equation (2) from equation (1):
\[ x - y = (t^2 + t + 1) - (t^2 - t + 1) \]
This simplifies to:
\[ x - y = 2t \]
Dividing the entire equation by 2 gives:
\[ t = \frac{x - y}{2} \quad (4) \]

### Step 4: Substitute \( t \) from equation (4) into equation (3)
Now, we will substitute \( t \) from equation (4) into equation (3):
\[ \frac{x + y}{2} = \left(\frac{x - y}{2}\right)^2 + 1 \]

### Step 5: Expand and simplify
Expanding the right side:
\[ \frac{x + y}{2} = \frac{(x - y)^2}{4} + 1 \]
Multiply through by 4 to eliminate the fractions:
\[ 2(x + y) = (x - y)^2 + 4 \]

### Step 6: Rearranging the equation
Rearranging gives:
\[ (x - y)^2 - 2(x + y) + 4 = 0 \]

### Step 7: Final Form
This is the equation of the locus. We can rewrite it as:
\[ x^2 + y^2 - 2xy - 2x - 2y + 4 = 0 \]

### Final Answer
The locus represented by the equations is:
\[ x^2 + y^2 - 2xy - 2x - 2y + 4 = 0 \]

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