For the parabola `y^2+4x-6y+13=0`, the vertex is .... force is ... directrix is ... L.R , is ……..

Answer

To solve the problem for the parabola given by the equation \( y^2 + 4x - 6y + 13 = 0 \), we will follow these steps:

### Step 1: Rearranging the Equation
Start by rearranging the equation into a standard form. We can group the \( y \) terms together and isolate \( x \):

\[ y^2 - 6y + 4x + 13 = 0 \]

### Step 2: Completing the Square
Next, we will complete the square for the \( y \) terms. The expression \( y^2 - 6y \) can be rewritten by adding and subtracting \( 9 \) (which is \( \left(\frac{6}{2}\right)^2 \)):

\[ (y^2 - 6y + 9) + 4x + 4 = 0 \]

This simplifies to:

\[ (y - 3)^2 + 4x + 4 = 0 \]

Rearranging gives us:

\[ (y - 3)^2 = -4(x + 1) \]

### Step 3: Identifying the Standard Form
Now, we can identify the standard form of the parabola. The equation \( (y - k)^2 = -4a(x - h) \) indicates that the vertex is at \( (h, k) \). Here, we have:

- \( h = -1 \)
- \( k = 3 \)
- \( 4a = 4 \) which gives \( a = 1 \)

### Step 4: Finding the Vertex
The vertex of the parabola is given by the coordinates \( (h, k) \):

\[ \text{Vertex} = (-1, 3) \]

### Step 5: Finding the Focus
The focus of the parabola is located at \( (h - a, k) \):

\[ \text{Focus} = (-1 - 1, 3) = (-2, 3) \]

### Step 6: Finding the Directrix
The directrix is a vertical line given by \( x = h + a \):

\[ \text{Directrix} = x = -1 + 1 = 0 \]

### Step 7: Finding the Length of the Latus Rectum
The length of the latus rectum is given by \( 4a \):

\[ \text{Length of Latus Rectum} = 4 \times 1 = 4 \]

### Summary of Results
- **Vertex**: \((-1, 3)\)
- **Focus**: \((-2, 3)\)
- **Directrix**: \(x = 0\)
- **Length of Latus Rectum**: \(4\)