The line `y=x+1` is 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 \( y = x + 1 \). The slope (m) of this line is 1.

### Step 2: Differentiate the curve
The curve is given by \( y^2 = 4x \). To find the slope of the tangent to the curve, we differentiate it implicitly with respect to \( x \).

Starting with:
\[ y^2 = 4x \]
Differentiating both sides:
\[ 2y \frac{dy}{dx} = 4 \]
Now, 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, the slope of the curve at the point of tangency must equal the slope of the line. Therefore, we set:
\[ \frac{2}{y_1} = 1 \]
Cross-multiplying gives:
\[ 2 = y_1 \]

### Step 4: Substitute \( y_1 \) back into the curve equation
Now that we have \( y_1 = 2 \), we substitute this value back into the curve equation to find \( x_1 \):
\[ y^2 = 4x \implies 2^2 = 4x \implies 4 = 4x \implies x = 1 \]
Thus, \( x_1 = 1 \).

### Step 5: Write the point of tangency
The point of tangency is \( (x_1, y_1) = (1, 2) \).

### Final Answer
The line \( y = x + 1 \) is tangent to the curve \( y^2 = 4x \) at the point \( (1, 2) \).

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