Find the centre, foci , the length of the axis, eccentricity and the equation of the directrices of the ellipse . `9x^2+16y^2-18x+32y-119=0`

Answer

To solve the problem step by step, we will follow the process of converting the given equation of the ellipse into standard form, then identify the required parameters such as center, foci, length of axes, eccentricity, and equations of the directrices.

### Step 1: Rewrite the given equation
The given equation is:
\[ 9x^2 + 16y^2 - 18x + 32y - 119 = 0 \]

### Step 2: Rearrange and group the terms
Group the \(x\) and \(y\) terms:
\[ 9x^2 - 18x + 16y^2 + 32y = 119 \]

### Step 3: Complete the square for \(x\) and \(y\)
For \(x\):
1. Factor out the coefficient of \(x^2\):
\[ 9(x^2 - 2x) \]
2. Complete the square:
\[ x^2 - 2x = (x-1)^2 - 1 \]
Thus,
\[ 9((x-1)^2 - 1) = 9(x-1)^2 - 9 \]

For \(y\):
1. Factor out the coefficient of \(y^2\):
\[ 16(y^2 + 2y) \]
2. Complete the square:
\[ y^2 + 2y = (y+1)^2 - 1 \]
Thus,
\[ 16((y+1)^2 - 1) = 16(y+1)^2 - 16 \]

### Step 4: Substitute back into the equation
Substituting back, we have:
\[ 9((x-1)^2 - 1) + 16((y+1)^2 - 1) = 119 \]
This simplifies to:
\[ 9(x-1)^2 + 16(y+1)^2 - 9 - 16 = 119 \]
\[ 9(x-1)^2 + 16(y+1)^2 = 144 \]

### Step 5: Divide by 144 to get standard form
\[ \frac{9(x-1)^2}{144} + \frac{16(y+1)^2}{144} = 1 \]
This simplifies to:
\[ \frac{(x-1)^2}{16} + \frac{(y+1)^2}{9} = 1 \]

### Step 6: Identify the parameters
From the equation \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\):
- Center \((h, k) = (1, -1)\)
- \(a^2 = 16 \Rightarrow a = 4\)
- \(b^2 = 9 \Rightarrow b = 3\)

### Step 7: Calculate the foci
The foci are given by:
\[ (h \pm c, k) \]
where \(c = \sqrt{a^2 - b^2} = \sqrt{16 - 9} = \sqrt{7}\).
Thus, the foci are:
\[ (1 \pm \sqrt{7}, -1) \]
So the foci are:
\[ (1 + \sqrt{7}, -1) \quad \text{and} \quad (1 - \sqrt{7}, -1) \]

### Step 8: Length of the axes
- Length of the major axis = \(2a = 2 \times 4 = 8\)
- Length of the minor axis = \(2b = 2 \times 3 = 6\)

### Step 9: Calculate the eccentricity
Eccentricity \(e\) is given by:
\[ e = \frac{c}{a} = \frac{\sqrt{7}}{4} \]

### Step 10: Find the equations of the directrices
The equations of the directrices are given by:
\[ x = h \pm \frac{a}{e} \]
Calculating:
\[ e = \frac{\sqrt{7}}{4} \Rightarrow \frac{a}{e} = \frac{4}{\frac{\sqrt{7}}{4}} = \frac{16}{\sqrt{7}} \]
Thus, the equations of the directrices are:
\[ x = 1 \pm \frac{16}{\sqrt{7}} \]

### Summary of Results
- **Center**: \((1, -1)\)
- **Foci**: \((1 + \sqrt{7}, -1)\) and \((1 - \sqrt{7}, -1)\)
- **Length of Major Axis**: \(8\)
- **Length of Minor Axis**: \(6\)
- **Eccentricity**: \(\frac{\sqrt{7}}{4}\)
- **Equations of Directrices**: \(x = 1 + \frac{16}{\sqrt{7}}\) and \(x = 1 - \frac{16}{\sqrt{7}}\)