The equation of the latus retum of the parabola `x^(2)+4x+2y=0` is

Options

A. `2y+1=0`
B. `2y=3`
C. `2y=-3`
D. `2y-1=0`

Answer

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

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

### Step 2: Complete the square for the \( x \) terms
To rewrite the equation in a more useful form, we complete the square for the \( x \) terms:
\[ y = -\frac{1}{2}(x^2 + 4x) \]
To complete the square, we take half of the coefficient of \( x \) (which is 4), square it, and add and subtract that value inside the parentheses:
\[ y = -\frac{1}{2}((x^2 + 4x + 4) - 4) \]
This simplifies to:
\[ y = -\frac{1}{2}((x + 2)^2 - 4) \]
\[ y = -\frac{1}{2}(x + 2)^2 + 2 \]

### Step 3: Identify the vertex and the parameter \( a \)
Now, we have the equation in the form:
\[ y = -\frac{1}{2}(x + 2)^2 + 2 \]
This indicates that the parabola opens downwards, and the vertex is at the point \( (-2, 2) \). The parameter \( a \) can be identified from the equation \( y = -\frac{1}{4a}(x - h)^2 + k \). Here, we have:
\[ -\frac{1}{4a} = -\frac{1}{2} \implies 4a = 2 \implies a = \frac{1}{2} \]

### Step 4: Write the equation of the latus rectum
The equation of the latus rectum for a parabola is given by:
\[ y = k \pm a \]
Since \( k = 2 \) and \( a = \frac{1}{2} \), we have:
\[ y = 2 \pm \frac{1}{2} \]
This gives us two equations:
\[ y = 2 + \frac{1}{2} = \frac{5}{2} \]
\[ y = 2 - \frac{1}{2} = \frac{3}{2} \]

### Step 5: Write the final equation
The latus rectum is a vertical line segment, and the equation can be expressed as:
\[ 2y - 5 = 0 \quad \text{and} \quad 2y - 3 = 0 \]

### Final Answer
Thus, the equation of the latus rectum of the parabola \( x^2 + 4x + 2y = 0 \) is:
\[ 2y - 1 = 0 \]