Find a point on the parabola `y=(x-2)^(2)`, where the tangent is parallel to the line joining (2, 0) and (4, 4).

Answer

To solve the problem of finding a point on the parabola \( y = (x - 2)^2 \) where the tangent is parallel to the line joining the points (2, 0) and (4, 4), we will follow these steps:

### Step 1: Find the slope of the line joining the points (2, 0) and (4, 4).
The slope \( m \) of a line through two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by the formula:
\[ m = \frac{y_2 - y_1}{x_2 - x_1} \]
Substituting the given points (2, 0) and (4, 4):
\[ m = \frac{4 - 0}{4 - 2} = \frac{4}{2} = 2 \]

### Step 2: Set the derivative of the parabola equal to the slope found in Step 1.
The slope of the tangent to the parabola at any point is given by the derivative \( \frac{dy}{dx} \). We first differentiate the function \( y = (x - 2)^2 \):
\[ \frac{dy}{dx} = 2(x - 2) \]
We want this slope to be equal to 2 (the slope of the line):
\[ 2(x - 2) = 2 \]

### Step 3: Solve for \( x \).
To find the value of \( x \), we can simplify the equation:
\[ x - 2 = 1 \quad \Rightarrow \quad x = 3 \]

### Step 4: Find the corresponding \( y \) value on the parabola.
Now that we have \( x = 3 \), we can substitute this back into the equation of the parabola to find \( y \):
\[ y = (3 - 2)^2 = 1^2 = 1 \]

### Step 5: Write the point.
Thus, the point on the parabola where the tangent is parallel to the line joining (2, 0) and (4, 4) is:
\[ (3, 1) \]

### Final Answer:
The point is \( (3, 1) \).

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