The equation of the tangent at the vertex of the parabola `x^(2)+4x+2y=0,` is

Options

A. x=-2
B. x=2
C. y=2
D. x=-a

Answer

To find the equation of the tangent at the vertex of the parabola given by the equation \( x^2 + 4x + 2y = 0 \), we will follow these steps:

### Step 1: Rewrite the equation
We start with the equation of the parabola:
\[ x^2 + 4x + 2y = 0 \]

### Step 2: Complete the square
To rewrite the quadratic in \( x \) in a more useful form, we complete the square:
\[ x^2 + 4x = (x + 2)^2 - 4 \]
Substituting this back into the equation gives:
\[ (x + 2)^2 - 4 + 2y = 0 \]
Rearranging this, we have:
\[ (x + 2)^2 = -2y + 4 \]
or
\[ (x + 2)^2 = -2(y - 2) \]

### Step 3: Identify the vertex
From the equation \( (x + 2)^2 = -2(y - 2) \), we can see that the vertex of the parabola is at the point \( (-2, 2) \).

### Step 4: Determine the equation of the tangent
For a parabola of the form \( (x - h)^2 = 4p(y - k) \), the tangent at the vertex (which is the point \( (h, k) \)) is a horizontal line. Since our vertex is at \( (-2, 2) \), the equation of the tangent line at this vertex is:
\[ y = k \]
where \( k = 2 \). Thus, the equation of the tangent is:
\[ y = 2 \]

### Final Answer
The equation of the tangent at the vertex of the parabola is:
\[ \boxed{y = 2} \]