Find the eccentricity of the hyperbola whose equation is `2x ^(2) - 3y ^(2) = 15.`

Answer

To find the eccentricity of the hyperbola given by the equation \(2x^2 - 3y^2 = 15\), we can follow these steps:

### Step 1: Rewrite the equation in standard form
We start with the equation:
\[ 2x^2 - 3y^2 = 15 \]
To convert this into standard form, we divide every term by 15:
\[ \frac{2x^2}{15} - \frac{3y^2}{15} = 1 \]
This simplifies to:
\[ \frac{x^2}{\frac{15}{2}} - \frac{y^2}{5} = 1 \]

### 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:
\[ a^2 = \frac{15}{2} \quad \text{and} \quad b^2 = 5 \]

### Step 3: Calculate \(c^2\)
The relationship between \(a\), \(b\), and \(c\) for hyperbolas is given by:
\[ c^2 = a^2 + b^2 \]
Substituting the values we found:
\[ c^2 = \frac{15}{2} + 5 \]
To add these, we convert 5 into a fraction with a denominator of 2:
\[ 5 = \frac{10}{2} \]
Thus,
\[ c^2 = \frac{15}{2} + \frac{10}{2} = \frac{25}{2} \]

### Step 4: Calculate \(c\)
Now, we take the square root to find \(c\):
\[ c = \sqrt{\frac{25}{2}} = \frac{5}{\sqrt{2}} \]

### Step 5: Calculate the eccentricity \(e\)
The eccentricity \(e\) of the hyperbola is given by:
\[ e = \frac{c}{a} \]
We need to find \(a\):
\[ a = \sqrt{a^2} = \sqrt{\frac{15}{2}} = \frac{\sqrt{15}}{\sqrt{2}} \]
Now substituting \(c\) and \(a\) into the formula for eccentricity:
\[ e = \frac{\frac{5}{\sqrt{2}}}{\frac{\sqrt{15}}{\sqrt{2}}} \]
The \(\sqrt{2}\) cancels out:
\[ e = \frac{5}{\sqrt{15}} \]

### Step 6: Simplify the eccentricity
To simplify \(\frac{5}{\sqrt{15}}\):
\[ e = \frac{5}{\sqrt{15}} = \frac{5}{\sqrt{3 \cdot 5}} = \frac{5}{\sqrt{5} \cdot \sqrt{3}} = \frac{5 \cdot \sqrt{5}}{5 \cdot \sqrt{3}} = \frac{\sqrt{5}}{\sqrt{3}} = \sqrt{\frac{5}{3}} \]

### Final Result
Thus, the eccentricity of the hyperbola is:
\[ e = \sqrt{\frac{5}{3}} \]