The line `y=x+1` is a tangent to the curve `y^(2)=4x` at the point :

Options

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

Answer

To find the point at which the line \( y = x + 1 \) is tangent to the curve \( y^2 = 4x \), we will follow these steps:

### Step 1: Identify the slope of the tangent line
The given line is \( y = x + 1 \). We can identify the slope \( m \) of the line:
\[ m = 1 \]

### Step 2: Differentiate the curve
The curve is given by the equation \( y^2 = 4x \). To find the slope of the tangent to the curve, we need to differentiate it implicitly:
\[ \frac{d}{dx}(y^2) = \frac{d}{dx}(4x) \]
Using the chain rule on the left side, we get:
\[ 2y \frac{dy}{dx} = 4 \]
Thus, we can solve for \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{4}{2y} = \frac{2}{y} \]

### Step 3: Set the slopes equal
Since the line is tangent to the curve at some point, the slopes must be equal. Therefore, we set the slope of the tangent line equal to the derivative of the curve:
\[ 1 = \frac{2}{y} \]

### Step 4: Solve for \( y \)
From the equation \( 1 = \frac{2}{y} \), we can solve for \( y \):
\[ y = 2 \]

### Step 5: Substitute \( y \) back into the curve equation
Now that we have \( y = 2 \), we can substitute this value back into the curve equation \( y^2 = 4x \) to find \( x \):
\[ 2^2 = 4x \implies 4 = 4x \implies x = 1 \]

### Step 6: Write the point of tangency
Thus, the point at which the line \( y = x + 1 \) is tangent to the curve \( y^2 = 4x \) is:
\[ (x, y) = (1, 2) \]

### Final Answer
The point of tangency is \( (1, 2) \).

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