Find the latus rectum, eccentricity, coordinates of the foci and the length of axes of the ellipse `4x^2 + 9y^2 - 8x - 36y+4=0`.

Answer

To solve the given problem step by step, we will start with the equation of the ellipse and manipulate it into standard form.

### Step 1: Write the equation of the ellipse
The given equation is:
\[ 4x^2 + 9y^2 - 8x - 36y + 4 = 0 \]

### Step 2: Rearrange the equation
We can rearrange the equation by grouping the \(x\) and \(y\) terms:
\[ 4x^2 - 8x + 9y^2 - 36y + 4 = 0 \]

### Step 3: Complete the square for \(x\) terms
Factor out the coefficient of \(x^2\):
\[ 4(x^2 - 2x) + 9y^2 - 36y + 4 = 0 \]
Now, complete the square for \(x\):
\[ x^2 - 2x \rightarrow (x - 1)^2 - 1 \]
Substituting back, we have:
\[ 4((x - 1)^2 - 1) + 9y^2 - 36y + 4 = 0 \]
This simplifies to:
\[ 4(x - 1)^2 - 4 + 9y^2 - 36y + 4 = 0 \]

### Step 4: Complete the square for \(y\) terms
Now, complete the square for \(y\):
\[ 9(y^2 - 4y) \rightarrow 9((y - 2)^2 - 4) \]
Substituting back, we have:
\[ 4(x - 1)^2 - 4 + 9((y - 2)^2 - 4) + 4 = 0 \]
This simplifies to:
\[ 4(x - 1)^2 + 9(y - 2)^2 - 36 = 0 \]

### Step 5: Set the equation to standard form
Rearranging gives:
\[ 4(x - 1)^2 + 9(y - 2)^2 = 36 \]
Dividing through by 36:
\[ \frac{(x - 1)^2}{9} + \frac{(y - 2)^2}{4} = 1 \]

### Step 6: Identify parameters of the ellipse
From the standard form \(\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1\), we have:
- Center \((h, k) = (1, 2)\)
- \(a^2 = 9 \Rightarrow a = 3\)
- \(b^2 = 4 \Rightarrow b = 2\)

### Step 7: Find the eccentricity
The eccentricity \(e\) 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 8: Find the coordinates of the foci
The foci are located at \((h \pm ae, k)\):
\[ Foci = (1 \pm 3 \cdot \frac{\sqrt{5}}{3}, 2) = (1 \pm \sqrt{5}, 2) \]
Thus, the coordinates of the foci are:
\[ (1 + \sqrt{5}, 2) \text{ and } (1 - \sqrt{5}, 2) \]

### Step 9: Find the lengths of the axes
- Length of the major axis = \(2a = 2 \times 3 = 6\)
- Length of the minor axis = \(2b = 2 \times 2 = 4\)

### Step 10: Find the length of the latus rectum
The length of the latus rectum \(L\) is given by:
\[ L = \frac{2b^2}{a} = \frac{2 \cdot 4}{3} = \frac{8}{3} \]

### Summary of Results
- Latus Rectum: \(\frac{8}{3}\)
- Eccentricity: \(\frac{\sqrt{5}}{3}\)
- Coordinates of the Foci: \((1 + \sqrt{5}, 2)\) and \((1 - \sqrt{5}, 2)\)
- Length of Major Axis: \(6\)
- Length of Minor Axis: \(4\)