Find vertex, focus, directrix and latus rectum of the parabola `y^(2)+4x+4y-3=0`.

Answer

To find the vertex, focus, directrix, and latus rectum of the parabola given by the equation \(y^2 + 4x + 4y - 3 = 0\), we can follow these steps:

### Step 1: Rearranging the equation
We start by rearranging the given equation into a standard form.

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

Adding 3 to both sides:

\[ y^2 + 4y + 4x = 3 \]

Next, we complete the square for the \(y\) terms.

### Step 2: Completing the square
The expression \(y^2 + 4y\) can be completed as follows:

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

Substituting this back into the equation gives:

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

This simplifies to:

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

\[ (y + 2)^2 + 4x = 7 \]

### Step 3: Isolating \(x\)
Now, we isolate \(x\):

\[ 4x = 7 - (y + 2)^2 \]

Dividing through by 4:

\[ x = \frac{7}{4} - \frac{(y + 2)^2}{4} \]

Rearranging gives us:

\[ (y + 2)^2 = -4\left(x - \frac{7}{4}\right) \]

### Step 4: Identifying the standard form
This is now in the standard form of a parabola that opens to the left:

\[ (y - k)^2 = -4a(x - h) \]

where \((h, k)\) is the vertex. From our equation, we can identify:

- \(h = \frac{7}{4}\)
- \(k = -2\)
- \(4a = 4 \Rightarrow a = 1\)

### Step 5: Finding the vertex
The vertex \((h, k)\) is:

\[ \text{Vertex} = \left(\frac{7}{4}, -2\right) \]

### Step 6: Finding the focus
The focus of the parabola is located at:

\[ (h - a, k) = \left(\frac{7}{4} - 1, -2\right) = \left(\frac{3}{4}, -2\right) \]

### Step 7: Finding the directrix
The equation of the directrix is given by:

\[ x = h + a = \frac{7}{4} + 1 = \frac{7}{4} + \frac{4}{4} = \frac{11}{4} \]

### Step 8: Finding the latus rectum
The length of the latus rectum is given by \(4a\). Since \(a = 1\):

\[ \text{Length of latus rectum} = 4 \times 1 = 4 \]

### Summary of results
- **Vertex**: \(\left(\frac{7}{4}, -2\right)\)
- **Focus**: \(\left(\frac{3}{4}, -2\right)\)
- **Directrix**: \(x = \frac{11}{4}\)
- **Latus Rectum**: Length = 4