The vertex of the parabola whose parametric equation is `x=t^2-t+1,y=t^2+t+1; t in R ,` is (1, 1) (b) (2, 2) `(1/2,1/2)` (d) `(3,3)`

Options

A. (1,1)
B. (2,2)
C. (1/2, 1/2)
D. (3,3)

Explanation

To find the vertex of the parabola given by the parametric equations \(x = t^2 - t + 1\) and \(y = t^2 + t + 1\), we can follow these steps:

### Step 1: Express \(t\) in terms of \(x\) and \(y\)
We start with the given parametric equations:
\[ x = t^2 - t + 1 \]
\[ y = t^2 + t + 1 \]

### Step 2: Combine the equations
We can add and subtract the equations to eliminate \(t\).

Adding the equations:
\[ x + y = (t^2 - t + 1) + (t^2 + t + 1) = 2t^2 + 2 \]
This simplifies to:
\[ x + y - 2 = 2t^2 \quad \text{(1)} \]

Subtracting the equations:
\[ y - x = (t^2 + t + 1) - (t^2 - t + 1) = 2t \]
This simplifies to:
\[ t = \frac{y - x}{2} \quad \text{(2)} \]

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

### Step 4: Rearrange the equation
Multiply through by 2 to eliminate the fraction:
\[ 2(x + y - 2) = (y - x)^2 \]
Expanding the left side:
\[ 2x + 2y - 4 = y^2 - 2xy + x^2 \]
Rearranging gives:
\[ x^2 - 2xy + y^2 - 2x - 2y + 4 = 0 \]

### Step 5: Identify the vertex
The vertex of the parabola can be found by determining the intersection of the axis of symmetry and the tangent at the vertex. The axis of symmetry can be derived from the quadratic form.

From the equation \(y - x = 0\) (which is the line \(y = x\)), we can substitute \(y = x\) into the rearranged equation:
\[ x^2 - 2x^2 + x^2 - 2x - 2x + 4 = 0 \]
This simplifies to:
\[ 0 = 0 \]
This means that we can find the vertex by substituting \(x = 1\) into either original parametric equation:
\[ x = 1 \implies y = 1 \]

### Conclusion
Thus, the vertex of the parabola is \((1, 1)\).