The eccentricity of the ellipse `9x^(2) + 5y^(2) - 18x - 20y - 16 = 0` is

Options

A. `1/2`
B. `2/3`
C. `1/3`
D. `3/4`

Answer

To find the eccentricity of the ellipse given by the equation \(9x^2 + 5y^2 - 18x - 20y - 16 = 0\), we will follow these steps:

### Step 1: Rearrange the equation
We start with the equation:
\[ 9x^2 + 5y^2 - 18x - 20y - 16 = 0 \]
Rearranging gives:
\[ 9x^2 - 18x + 5y^2 - 20y = 16 \]

### Step 2: Complete the square for \(x\) and \(y\)
For the \(x\) terms:
\[ 9(x^2 - 2x) \quad \text{(factor out 9)} \]
To complete the square:
\[ x^2 - 2x = (x - 1)^2 - 1 \]
Thus:
\[ 9((x - 1)^2 - 1) = 9(x - 1)^2 - 9 \]

For the \(y\) terms:
\[ 5(y^2 - 4y) \quad \text{(factor out 5)} \]
To complete the square:
\[ y^2 - 4y = (y - 2)^2 - 4 \]
Thus:
\[ 5((y - 2)^2 - 4) = 5(y - 2)^2 - 20 \]

### Step 3: Substitute back into the equation
Substituting back into the rearranged equation:
\[ 9((x - 1)^2 - 1) + 5((y - 2)^2 - 4) = 16 \]
This simplifies to:
\[ 9(x - 1)^2 - 9 + 5(y - 2)^2 - 20 = 16 \]
Combining like terms:
\[ 9(x - 1)^2 + 5(y - 2)^2 - 29 = 16 \]
Thus:
\[ 9(x - 1)^2 + 5(y - 2)^2 = 45 \]

### Step 4: Divide by 45 to get the standard form
Dividing through by 45 gives:
\[ \frac{(x - 1)^2}{5} + \frac{(y - 2)^2}{9} = 1 \]

### Step 5: Identify \(a^2\) and \(b^2\)
From the standard form \(\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1\):
- \(a^2 = 5\)
- \(b^2 = 9\)

### Step 6: Calculate the eccentricity
The eccentricity \(e\) of an ellipse is given by the formula:
\[ e = \sqrt{1 - \frac{a^2}{b^2}} \]
Since \(b^2 > a^2\), we have:
\[ e = \sqrt{1 - \frac{5}{9}} = \sqrt{\frac{4}{9}} = \frac{2}{3} \]

### Final Answer
The eccentricity of the ellipse is:
\[ \frac{2}{3} \]

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