Equation of tangent at the vertex of parabola `x^(2)+8x+4y=0` is

Options

A. `x=4`
B. `x=-4`
C. `y=4`
D. `y=-4`

Answer

To find the equation of the tangent at the vertex of the parabola given by the equation \(x^2 + 8x + 4y = 0\), we can follow these steps:

### Step 1: Rewrite the equation in standard form
We start with the equation of the parabola:
\[ x^2 + 8x + 4y = 0 \]
We can rearrange this to isolate \(y\):
\[ 4y = -x^2 - 8x \]
\[ y = -\frac{1}{4}(x^2 + 8x) \]

### Step 2: Complete the square
Next, we complete the square for the expression in parentheses:
\[ x^2 + 8x = (x + 4)^2 - 16 \]
Substituting this back into the equation for \(y\):
\[ y = -\frac{1}{4}((x + 4)^2 - 16) \]
\[ y = -\frac{1}{4}(x + 4)^2 + 4 \]

### Step 3: Identify the vertex
From the equation \(y = -\frac{1}{4}(x + 4)^2 + 4\), we can see that the vertex of the parabola is at the point \((-4, 4)\).

### Step 4: Find the equation of the tangent at the vertex
The tangent line at the vertex of a parabola in the form \(y = ax^2 + bx + c\) is horizontal if the parabola opens up or down. Since our parabola opens downwards (as indicated by the negative coefficient of the quadratic term), the tangent line at the vertex will be horizontal.

The equation of a horizontal line passing through the vertex \((-4, 4)\) is:
\[ y = 4 \]

### Final Answer
Thus, the equation of the tangent at the vertex of the parabola \(x^2 + 8x + 4y = 0\) is:
\[ \boxed{y = 4} \]