The tangents to the curve `y = (x-2)^(2)-1` at its points of intersection with the line `x - y = 3`, intersect at the point:

Options

A. `((5)/(2),1)`
B. `(-(5)/(2),-1)`
C. `((5)/(2),-1)`
D. `(-(5)/(2),1)`

Answer

To solve the problem, we need to find the point where the tangents to the curve \( y = (x - 2)^2 - 1 \) at its points of intersection with the line \( x - y = 3 \) intersect.

### Step 1: Find the points of intersection of the curve and the line

First, we need to express the line \( x - y = 3 \) in terms of \( y \):
\[ y = x - 3 \]

Next, we substitute this expression for \( y \) into the equation of the curve:
\[ x - 3 = (x - 2)^2 - 1 \]

Expanding the right-hand side:
\[ x - 3 = (x^2 - 4x + 4) - 1 \]
\[ x - 3 = x^2 - 4x + 3 \]

Rearranging the equation gives us:
\[ 0 = x^2 - 5x + 6 \]

### Step 2: Solve the quadratic equation

Now, we can factor the quadratic:
\[ 0 = (x - 2)(x - 3) \]

Thus, the solutions for \( x \) are:
\[ x = 2 \quad \text{and} \quad x = 3 \]

### Step 3: Find the corresponding \( y \) values

Now, we substitute these \( x \) values back into the equation of the line to find the corresponding \( y \) values.

For \( x = 2 \):
\[ y = 2 - 3 = -1 \quad \Rightarrow \quad (2, -1) \]

For \( x = 3 \):
\[ y = 3 - 3 = 0 \quad \Rightarrow \quad (3, 0) \]

So, the points of intersection are \( A(2, -1) \) and \( B(3, 0) \).

### Step 4: Find the equations of the tangents at points A and B

The derivative of the curve \( y = (x - 2)^2 - 1 \) is:
\[ \frac{dy}{dx} = 2(x - 2) \]

Calculating the slope at point \( A(2, -1) \):
\[ \text{slope at } A = 2(2 - 2) = 0 \]
The equation of the tangent at \( A \) is:
\[ y + 1 = 0 \quad \Rightarrow \quad y = -1 \]

Calculating the slope at point \( B(3, 0) \):
\[ \text{slope at } B = 2(3 - 2) = 2 \]
The equation of the tangent at \( B \) is:
\[ y - 0 = 2(x - 3) \quad \Rightarrow \quad y = 2x - 6 \]

### Step 5: Find the intersection of the two tangents

Now we need to find the intersection of the lines \( y = -1 \) and \( y = 2x - 6 \).

Setting the equations equal to each other:
\[ -1 = 2x - 6 \]
Solving for \( x \):
\[ 2x = 5 \quad \Rightarrow \quad x = \frac{5}{2} \]

Substituting \( x = \frac{5}{2} \) back into \( y = -1 \):
\[ y = -1 \]

### Final Result

Thus, the point where the tangents intersect is:
\[ \left( \frac{5}{2}, -1 \right) \]