If the line x-y+k =0 is a tangent to the parabola `x^(2) = 4y` then k = ?

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the problem of finding the value of \( k \) such that the line \( x - y + k = 0 \) is a tangent to the parabola \( x^2 = 4y \), we can follow these steps:

### Step 1: Identify the parabola and its parameters
The given parabola is \( x^2 = 4y \). We can rewrite it in the standard form \( y = \frac{1}{4}x^2 \).

### Step 2: Identify the line and express it in slope-intercept form
The line given is \( x - y + k = 0 \). Rearranging this, we get:
\[ y = x + k \]
This shows that the slope \( m \) of the line is \( 1 \) and the y-intercept \( c \) is \( k \).

### Step 3: Use the condition for tangency
For the line to be a tangent to the parabola, the following condition must hold:
\[ c = \frac{a}{m} \]
where \( a \) is the parameter from the parabola's equation \( x^2 = 4ay \). Here, \( a = 1 \) (since \( 4a = 4 \)). The slope \( m = 1 \) (from the line).

### Step 4: Substitute the values into the condition
Substituting \( a \) and \( m \) into the tangency condition:
\[ k = \frac{1}{1} \]
This simplifies to:
\[ k = 1 \]

### Conclusion
Thus, the value of \( k \) such that the line \( x - y + k = 0 \) is a tangent to the parabola \( x^2 = 4y \) is:
\[ \boxed{1} \]