The eccentricity of the conic `9x^(2) + 25y^(2) - 18 x - 100 y = 116` is

Options

A. A. `(3)/(5)`
B. B. `(16)/(25)`
C. C. `(4)/(5)`
D. D. `(9)/(25)`

Answer

To find the eccentricity of the conic given by the equation \(9x^2 + 25y^2 - 18x - 100y = 116\), we will follow these steps:

### Step 1: Rearrange the equation
Start by moving all terms to one side of the equation:
\[ 9x^2 + 25y^2 - 18x - 100y - 116 = 0 \]

### Step 2: Group the terms
Group the \(x\) terms and the \(y\) terms:
\[ (9x^2 - 18x) + (25y^2 - 100y) = 116 \]

### Step 3: Complete the square for \(x\)
For the \(x\) terms \(9x^2 - 18x\):
1. Factor out 9:
\[ 9(x^2 - 2x) \]
2. Complete the square:
\[ x^2 - 2x = (x-1)^2 - 1 \]
Thus,
\[ 9((x-1)^2 - 1) = 9(x-1)^2 - 9 \]

### Step 4: Complete the square for \(y\)
For the \(y\) terms \(25y^2 - 100y\):
1. Factor out 25:
\[ 25(y^2 - 4y) \]
2. Complete the square:
\[ y^2 - 4y = (y-2)^2 - 4 \]
Thus,
\[ 25((y-2)^2 - 4) = 25(y-2)^2 - 100 \]

### Step 5: Substitute back into the equation
Substituting the completed squares back into the equation gives:
\[ 9((x-1)^2 - 1) + 25((y-2)^2 - 4) = 116 \]
This simplifies to:
\[ 9(x-1)^2 - 9 + 25(y-2)^2 - 100 = 116 \]
Combining constant terms:
\[ 9(x-1)^2 + 25(y-2)^2 - 109 = 116 \]
Thus,
\[ 9(x-1)^2 + 25(y-2)^2 = 225 \]

### Step 6: Divide through by 225
To get the standard form of the ellipse:
\[ \frac{9(x-1)^2}{225} + \frac{25(y-2)^2}{225} = 1 \]
This simplifies to:
\[ \frac{(x-1)^2}{25} + \frac{(y-2)^2}{9} = 1 \]

### Step 7: Identify \(a^2\) and \(b^2\)
From the standard form \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\):
- \(a^2 = 25\) (hence \(a = 5\))
- \(b^2 = 9\) (hence \(b = 3\))

### Step 8: Calculate the eccentricity \(e\)
The eccentricity \(e\) of an ellipse is given by:
\[ e = \sqrt{1 - \frac{b^2}{a^2}} \]
Substituting the values:
\[ e = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{25 - 9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5} \]

### Final Answer
Thus, the eccentricity of the conic is:
\[ \boxed{\frac{4}{5}} \]