If `y=2x+k` is a tangent to the curve `x^(2)=4y`, then k is equal to

Options

A. 4
B. 43467
C. -4
D. `-1//2`

Answer

To find the value of \( k \) such that the line \( y = 2x + k \) is a tangent to the curve \( x^2 = 4y \), we can follow these steps:

### Step 1: Substitute the equation of the line into the curve equation.
We start with the curve equation:
\[ x^2 = 4y \]
Substituting \( y = 2x + k \) into the curve equation gives:
\[ x^2 = 4(2x + k) \]

### Step 2: Simplify the equation.
Expanding the right side:
\[ x^2 = 8x + 4k \]
Rearranging this equation leads to:
\[ x^2 - 8x - 4k = 0 \]

### Step 3: Use the condition for tangency.
For the line to be a tangent to the curve, the quadratic equation must have equal roots. This occurs when the discriminant \( D \) is equal to zero. The discriminant for a quadratic equation \( ax^2 + bx + c = 0 \) is given by:
\[ D = b^2 - 4ac \]
In our case, \( a = 1 \), \( b = -8 \), and \( c = -4k \). Therefore, the discriminant is:
\[ D = (-8)^2 - 4(1)(-4k) = 64 + 16k \]
Setting the discriminant equal to zero for tangency:
\[ 64 + 16k = 0 \]

### Step 4: Solve for \( k \).
Now, we can solve for \( k \):
\[ 16k = -64 \]
\[ k = \frac{-64}{16} = -4 \]

### Conclusion:
Thus, the value of \( k \) is:
\[ \boxed{-4} \]

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