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

Answer

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

### Step 1: Rearranging the Equation
Start with the given equation:
\[ 4x^2 + 16y^2 - 24x - 32y = 12 \]
Rearranging gives:
\[ 4x^2 - 24x + 16y^2 - 32y = 12 \]

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

**For \(x\):**
\[ 4(x^2 - 6x) = 4\left((x - 3)^2 - 9\right) = 4(x - 3)^2 - 36 \]

**For \(y\):**
\[ 16(y^2 - 2y) = 16\left((y - 1)^2 - 1\right) = 16(y - 1)^2 - 16 \]

Substituting back into the equation gives:
\[ 4(x - 3)^2 - 36 + 16(y - 1)^2 - 16 = 12 \]
Simplifying this:
\[ 4(x - 3)^2 + 16(y - 1)^2 - 52 = 12 \]
\[ 4(x - 3)^2 + 16(y - 1)^2 = 64 \]

### Step 3: Dividing by 64
To get the standard form of the ellipse, divide the entire equation by 64:
\[ \frac{(x - 3)^2}{16} + \frac{(y - 1)^2}{4} = 1 \]

### Step 4: Identifying the Parameters
From the standard form \(\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1\), we identify:
- Center \((h, k) = (3, 1)\)
- \(a^2 = 16 \Rightarrow a = 4\)
- \(b^2 = 4 \Rightarrow b = 2\)

### Step 5: Finding Vertices
The vertices of the ellipse are located at:
- Along the major axis (horizontal):
\((h \pm a, k) = (3 \pm 4, 1) = (7, 1) \text{ and } (-1, 1)\)
- Along the minor axis (vertical):
\((h, k \pm b) = (3, 1 \pm 2) = (3, 3) \text{ and } (3, -1)\)

### Step 6: Finding Foci
The distance to the foci \(c\) is calculated using \(c^2 = a^2 - b^2\):
\[ c^2 = 16 - 4 = 12 \Rightarrow c = 2\sqrt{3} \]
The foci are located at:
\((h \pm c, k) = (3 \pm 2\sqrt{3}, 1)\)

### Step 7: Finding Eccentricity
The eccentricity \(e\) is given by:
\[ e = \frac{c}{a} = \frac{2\sqrt{3}}{4} = \frac{\sqrt{3}}{2} \]

### Step 8: Finding Directrices
The equations of the directrices are:
\[ x = h \pm \frac{a}{e} = 3 \pm \frac{4}{\frac{\sqrt{3}}{2}} = 3 \pm \frac{8}{\sqrt{3}} \]

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