Show that `4x^(2) + 16y^(2) - 24x - 32y - 12 = 0 ` is equation of an ellipse : and find its foci, vertices, eccentricity and directrices.

Answer

To show that the equation \(4x^2 + 16y^2 - 24x - 32y - 12 = 0\) represents an ellipse and to find its foci, vertices, eccentricity, and directrices, we will follow these steps:

### Step 1: Rearranging the Equation
Start with the given equation:
\[ 4x^2 + 16y^2 - 24x - 32y - 12 = 0 \]
Divide the entire equation by 4 to simplify:
\[ x^2 + 4y^2 - 6x - 8y - 3 = 0 \]

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

1. For \(x\):
\[ x^2 - 6x \quad \text{(take half of -6, square it: } (-3)^2 = 9\text{)} \]
This can be rewritten as:
\[ (x - 3)^2 - 9 \]

2. For \(y\):
\[ 4y^2 - 8y \quad \text{(factor out 4: } 4(y^2 - 2y)\text{)} \]
Now complete the square inside the parentheses:
\[ y^2 - 2y \quad \text{(take half of -2, square it: } (-1)^2 = 1\text{)} \]
This can be rewritten as:
\[ 4((y - 1)^2 - 1) = 4(y - 1)^2 - 4 \]

Putting it all together:
\[ (x - 3)^2 - 9 + 4(y - 1)^2 - 4 - 3 = 0 \]
This simplifies to:
\[ (x - 3)^2 + 4(y - 1)^2 - 16 = 0 \]
or:
\[ (x - 3)^2 + 4(y - 1)^2 = 16 \]

### Step 3: Standard Form of the Ellipse
Now, divide the entire equation by 16:
\[ \frac{(x - 3)^2}{16} + \frac{(y - 1)^2}{4} = 1 \]
This is in the standard form of an ellipse:
\[ \frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1 \]
where \(h = 3\), \(k = 1\), \(a^2 = 16\) (thus \(a = 4\)), and \(b^2 = 4\) (thus \(b = 2\)).

### Step 4: Identifying the Components
1. **Center**: The center of the ellipse is at \((h, k) = (3, 1)\).
2. **Vertices**: The vertices are at:
\[ (h \pm a, k) = (3 \pm 4, 1) \Rightarrow (7, 1) \text{ and } (-1, 1) \]
3. **Foci**: The distance of the foci from the center is given by \(c = \sqrt{a^2 - b^2} = \sqrt{16 - 4} = \sqrt{12} = 2\sqrt{3}\). Thus, the foci are at:
\[ (h \pm c, k) = (3 \pm 2\sqrt{3}, 1) \Rightarrow (3 + 2\sqrt{3}, 1) \text{ and } (3 - 2\sqrt{3}, 1) \]
4. **Eccentricity**: The eccentricity \(e\) is given by:
\[ e = \frac{c}{a} = \frac{2\sqrt{3}}{4} = \frac{\sqrt{3}}{2} \]
5. **Directrices**: The equations of the directrices are given by:
\[ x = h \pm \frac{a}{e} = 3 \pm \frac{4}{\frac{\sqrt{3}}{2}} = 3 \pm \frac{8}{\sqrt{3}} = 3 \pm \frac{8\sqrt{3}}{3} \]

### Summary of Results
- **Center**: \((3, 1)\)
- **Vertices**: \((7, 1)\) and \((-1, 1)\)
- **Foci**: \((3 + 2\sqrt{3}, 1)\) and \((3 - 2\sqrt{3}, 1)\)
- **Eccentricity**: \(\frac{\sqrt{3}}{2}\)
- **Directrices**: \(x = 3 \pm \frac{8\sqrt{3}}{3}\)