Find the latus rectum, eccentricity and foci of the curve `4x^(2)+9y^(2)-8x-36y+4=0`

Answer

To solve the equation \(4x^2 + 9y^2 - 8x - 36y + 4 = 0\) and find the latus rectum, eccentricity, and foci of the ellipse, we will follow these steps:

### Step 1: Rearranging the Equation
We start with the given equation:
\[ 4x^2 + 9y^2 - 8x - 36y + 4 = 0 \]
We rearrange it to group the \(x\) and \(y\) terms:
\[ 4x^2 - 8x + 9y^2 - 36y + 4 = 0 \]

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

**For \(x\):**
\[ 4(x^2 - 2x) \quad \text{(factor out the 4)} \]
To complete the square:
\[ x^2 - 2x = (x - 1)^2 - 1 \]
Thus,
\[ 4((x - 1)^2 - 1) = 4(x - 1)^2 - 4 \]

**For \(y\):**
\[ 9(y^2 - 4y) \quad \text{(factor out the 9)} \]
To complete the square:
\[ y^2 - 4y = (y - 2)^2 - 4 \]
Thus,
\[ 9((y - 2)^2 - 4) = 9(y - 2)^2 - 36 \]

### Step 3: Substitute Back into the Equation
Substituting back into the equation gives:
\[ 4(x - 1)^2 - 4 + 9(y - 2)^2 - 36 + 4 = 0 \]
Simplifying this:
\[ 4(x - 1)^2 + 9(y - 2)^2 - 36 = 0 \]
\[ 4(x - 1)^2 + 9(y - 2)^2 = 36 \]

### Step 4: Divide by 36
Now, we divide the entire equation by 36 to get it into standard form:
\[ \frac{(x - 1)^2}{9} + \frac{(y - 2)^2}{4} = 1 \]

### Step 5: Identify Parameters
From the standard form \(\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1\), we identify:
- \(h = 1\), \(k = 2\)
- \(a^2 = 9 \Rightarrow a = 3\)
- \(b^2 = 4 \Rightarrow b = 2\)

### Step 6: Find the Eccentricity
The eccentricity \(e\) of the ellipse is given by:
\[ e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{4}{9}} = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3} \]

### Step 7: Find the Foci
The foci of the ellipse are located at:
\[ (h \pm ae, k) = (1 \pm 3e, 2) \]
Calculating the foci:
\[ F_1 = \left(1 + 3 \cdot \frac{\sqrt{5}}{3}, 2\right) = (1 + \sqrt{5}, 2) \]
\[ F_2 = \left(1 - 3 \cdot \frac{\sqrt{5}}{3}, 2\right) = (1 - \sqrt{5}, 2) \]

### Step 8: Find the Latus Rectum
The latus rectum \(L\) of the ellipse is given by:
\[ L = \frac{2b^2}{a} = \frac{2 \cdot 4}{3} = \frac{8}{3} \]

### Final Results
- **Latus Rectum**: \(\frac{8}{3}\)
- **Eccentricity**: \(\frac{\sqrt{5}}{3}\)
- **Foci**: \((1 + \sqrt{5}, 2)\) and \((1 - \sqrt{5}, 2)\)