The minimum possible distnace between the points `A(a, a-1)` and `B(b, b^(2)+b+1)AA a, b in R` is D units, then the value of `D^(2)` is

Answer

To find the minimum possible distance between the points \( A(a, a-1) \) and \( B(b, b^2 + b + 1) \), we will follow these steps:

### Step 1: Understand the Points
The coordinates of point \( A \) are given as \( A(a, a-1) \) and those of point \( B \) as \( B(b, b^2 + b + 1) \).

### Step 2: Identify the Locus of Points
- The point \( A \) lies on the line \( y = x - 1 \).
- The point \( B \) lies on the parabola \( y = x^2 + x + 1 \).

### Step 3: Find the Distance Formula
The distance \( D \) between points \( A \) and \( B \) can be expressed as:
\[ D = \sqrt{(b - a)^2 + ((b^2 + b + 1) - (a - 1))^2} \]
This simplifies to:
\[ D = \sqrt{(b - a)^2 + (b^2 + b - a + 2)^2} \]

### Step 4: Minimize the Distance
To find the minimum distance, we can use calculus. However, a geometric interpretation can simplify the process. The minimum distance occurs when the line connecting points \( A \) and \( B \) is perpendicular to the tangent of the parabola at point \( B \).

### Step 5: Find the Slope of the Line and Parabola
1. The slope of the line \( y = x - 1 \) is \( 1 \).
2. The derivative of the parabola \( y = x^2 + x + 1 \) gives the slope:
\[ \frac{dy}{dx} = 2b + 1 \]
We set this equal to \( 1 \) to find the point of tangency:
\[ 2b + 1 = 1 \implies 2b = 0 \implies b = 0 \]

### Step 6: Find the Corresponding Point on the Parabola
Substituting \( b = 0 \) into the parabola:
\[ y = 0^2 + 0 + 1 = 1 \]
Thus, the point \( B \) is \( (0, 1) \).

### Step 7: Find the Corresponding Point on the Line
Substituting \( a = 0 \) into the line:
\[ y = 0 - 1 = -1 \]
Thus, the point \( A \) is \( (0, -1) \).

### Step 8: Calculate the Minimum Distance
Now, we can calculate the distance between points \( A(0, -1) \) and \( B(0, 1) \):
\[ D = \sqrt{(0 - 0)^2 + (1 - (-1))^2} = \sqrt{0 + (1 + 1)^2} = \sqrt{4} = 2 \]

### Step 9: Find \( D^2 \)
Finally, we find \( D^2 \):
\[ D^2 = 2^2 = 4 \]

Thus, the value of \( D^2 \) is \( \boxed{4} \).