Show that `4x^(2) + 8x + y^(2) - 4y + 4 = 0` represents an ellipse. Find its eccentricity, co-ordinates of foci equations of major and minor axes and latus - rectum.

Answer

To solve the equation \(4x^2 + 8x + y^2 - 4y + 4 = 0\) and show that it represents an ellipse, we will follow these steps:

### Step 1: Rearranging the Equation
We start by rearranging the given equation:
\[ 4x^2 + 8x + y^2 - 4y + 4 = 0 \]
This can be rewritten as:
\[ 4x^2 + 8x + y^2 - 4y = -4 \]

### Step 2: Completing the Square
Next, we will complete the square for the \(x\) and \(y\) terms.

**For \(x\):**
\[ 4(x^2 + 2x) = 4((x + 1)^2 - 1) = 4(x + 1)^2 - 4 \]

**For \(y\):**
\[ y^2 - 4y = (y - 2)^2 - 4 \]

Substituting these back into the equation gives:
\[ 4((x + 1)^2 - 1) + (y - 2)^2 - 4 = -4 \]
This simplifies to:
\[ 4(x + 1)^2 + (y - 2)^2 - 4 - 4 = -4 \]
Thus:
\[ 4(x + 1)^2 + (y - 2)^2 = 4 \]

### Step 3: Dividing by 4
Now, we divide the entire equation by 4:
\[ \frac{(x + 1)^2}{1} + \frac{(y - 2)^2}{4} = 1 \]

### Step 4: Identifying the Form of the Ellipse
This is in the standard form of an ellipse:
\[ \frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1 \]
where \((h, k)\) is the center, \(a^2 = 1\), and \(b^2 = 4\).

### Step 5: Finding the Center, \(a\), and \(b\)
From the equation:
- Center \((h, k) = (-1, 2)\)
- \(a = 1\) (semi-major axis)
- \(b = 2\) (semi-minor axis)

### Step 6: Finding the Eccentricity
The eccentricity \(e\) of an ellipse is given by:
\[ e = \sqrt{1 - \frac{a^2}{b^2}} = \sqrt{1 - \frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \]

### Step 7: Finding the Coordinates of the Foci
The foci of the ellipse are located at:
\[ (h, k \pm c) \quad \text{where } c = \sqrt{b^2 - a^2} = \sqrt{4 - 1} = \sqrt{3} \]
Thus, the coordinates of the foci are:
\[ (-1, 2 + \sqrt{3}) \quad \text{and} \quad (-1, 2 - \sqrt{3}) \]

### Step 8: Equations of the Major and Minor Axes
- The major axis is vertical (along the y-axis):
\[ x = -1 \]
- The minor axis is horizontal (along the x-axis):
\[ y = 2 \]

### Step 9: Finding the Length of the Latus Rectum
The length of the latus rectum \(L\) is given by:
\[ L = \frac{2b^2}{a} = \frac{2 \cdot 4}{1} = 8 \]

### Step 10: Finding the Equations of the Latus Rectum
The equations of the latus rectum can be found using the foci:
For the focus at \((-1, 2 + \sqrt{3})\):
\[ y - (2 + \sqrt{3}) = \pm \frac{b}{a}(x + 1) \quad \Rightarrow \quad y - (2 + \sqrt{3}) = \pm 2(x + 1) \]
For the focus at \((-1, 2 - \sqrt{3})\):
\[ y - (2 - \sqrt{3}) = \pm 2(x + 1) \]

### Summary of Results
- **Eccentricity**: \(e = \frac{\sqrt{3}}{2}\)
- **Coordinates of Foci**: \((-1, 2 + \sqrt{3})\) and \((-1, 2 - \sqrt{3})\)
- **Equation of Major Axis**: \(x = -1\)
- **Equation of Minor Axis**: \(y = 2\)
- **Length of Latus Rectum**: \(8\)