Find the centre, the length of axes, the eccentricity and the foci of the ellipse `12x^(2)+4y^(2)+24x-16y+25=0`

Answer

To solve the problem of finding the center, length of axes, eccentricity, and foci of the ellipse given by the equation \(12x^2 + 4y^2 + 24x - 16y + 25 = 0\), we will follow these steps:

### Step 1: Rearranging the Equation
First, we need to rearrange the equation into the standard form of an ellipse. We start with:
\[ 12x^2 + 4y^2 + 24x - 16y + 25 = 0 \]
We can rearrange it as:
\[ 12x^2 + 24x + 4y^2 - 16y + 25 = 0 \]

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

**For \(x\):**
\[ 12(x^2 + 2x) = 12((x+1)^2 - 1) = 12(x+1)^2 - 12 \]

**For \(y\):**
\[ 4(y^2 - 4y) = 4((y-2)^2 - 4) = 4(y-2)^2 - 16 \]

Now substituting back into the equation:
\[ 12((x+1)^2 - 1) + 4((y-2)^2 - 4) + 25 = 0 \]
This simplifies to:
\[ 12(x+1)^2 - 12 + 4(y-2)^2 - 16 + 25 = 0 \]
\[ 12(x+1)^2 + 4(y-2)^2 - 3 = 0 \]
\[ 12(x+1)^2 + 4(y-2)^2 = 3 \]

### Step 3: Dividing by 3
To get the standard form, divide the entire equation by 3:
\[ \frac{12(x+1)^2}{3} + \frac{4(y-2)^2}{3} = 1 \]
This simplifies to:
\[ \frac{4(x+1)^2}{1} + \frac{(y-2)^2}{\frac{3}{4}} = 1 \]

### Step 4: Identifying Parameters
From the equation \(\frac{(x+1)^2}{\frac{1}{4}} + \frac{(y-2)^2}{\frac{3}{4}} = 1\), we can identify:
- Center: \((-1, 2)\)
- \(a^2 = \frac{1}{4} \Rightarrow a = \frac{1}{2}\)
- \(b^2 = \frac{3}{4} \Rightarrow b = \frac{\sqrt{3}}{2}\)

### Step 5: Length of Axes
- Length of the major axis (along the \(y\)-axis): \(2b = 2 \cdot \frac{\sqrt{3}}{2} = \sqrt{3}\)
- Length of the minor axis (along the \(x\)-axis): \(2a = 2 \cdot \frac{1}{2} = 1\)

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

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

### Summary of Results
- Center: \((-1, 2)\)
- Length of Major Axis: \(\sqrt{3}\)
- Length of Minor Axis: \(1\)
- Eccentricity: \(\frac{\sqrt{6}}{3}\)
- Foci: \(\left(-1, 2 + \frac{1}{\sqrt{2}}\right)\) and \(\left(-1, 2 - \frac{1}{\sqrt{2}}\right)\)