Find the eccentricity and the coordinate of foci of the hyperbola `25 x ^(2) - 9y ^(2) = 225.`

Answer

To find the eccentricity and the coordinates of the foci of the hyperbola given by the equation \( 25x^2 - 9y^2 = 225 \), we will follow these steps:

### Step 1: Rewrite the equation in standard form
We start with the equation of the hyperbola:
\[ 25x^2 - 9y^2 = 225 \]
To convert this into standard form, we divide the entire equation by 225:
\[ \frac{25x^2}{225} - \frac{9y^2}{225} = 1 \]
This simplifies to:
\[ \frac{x^2}{9} - \frac{y^2}{25} = 1 \]

### Step 2: Identify \(a^2\) and \(b^2\)
From the standard form \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), we can identify:
\[ a^2 = 9 \quad \text{and} \quad b^2 = 25 \]
Thus, we find:
\[ a = 3 \quad \text{and} \quad b = 5 \]

### Step 3: Calculate the eccentricity \(e\)
The eccentricity \(e\) of a hyperbola is given by the formula:
\[ e = \sqrt{1 + \frac{b^2}{a^2}} \]
Substituting the values of \(b^2\) and \(a^2\):
\[ e = \sqrt{1 + \frac{25}{9}} = \sqrt{1 + \frac{25}{9}} = \sqrt{\frac{34}{9}} = \frac{\sqrt{34}}{3} \]

### Step 4: Find the coordinates of the foci
The coordinates of the foci for a hyperbola are given by \((\pm ae, 0)\). We already have \(a = 3\) and \(e = \frac{\sqrt{34}}{3}\). Therefore, we calculate:
\[ ae = 3 \cdot \frac{\sqrt{34}}{3} = \sqrt{34} \]
Thus, the coordinates of the foci are:
\[ (\sqrt{34}, 0) \quad \text{and} \quad (-\sqrt{34}, 0) \]

### Final Answer
- Eccentricity \(e = \frac{\sqrt{34}}{3}\)
- Coordinates of the foci: \((\sqrt{34}, 0)\) and \((- \sqrt{34}, 0)\)

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