If `5/4` is the eccentricity of a hyperbola find the eccentricity of its conjugate hyperbola.

Answer

To find the eccentricity of the conjugate hyperbola given that the eccentricity of the hyperbola is \( \frac{5}{4} \), we can follow these steps:

### Step 1: Understand the relationship between the eccentricities
The relationship between the eccentricity of a hyperbola \( E_2 \) and its conjugate hyperbola \( E_1 \) is given by the formula:
\[ \frac{1}{E_1^2} + \frac{1}{E_2^2} = 1 \]

### Step 2: Assign the known value
Let \( E_2 = \frac{5}{4} \). We need to find \( E_1 \).

### Step 3: Substitute the known value into the equation
Substituting \( E_2 \) into the relationship:
\[ \frac{1}{E_1^2} + \frac{1}{\left(\frac{5}{4}\right)^2} = 1 \]

### Step 4: Calculate \( \left(\frac{5}{4}\right)^2 \)
Calculating \( \left(\frac{5}{4}\right)^2 \):
\[ \left(\frac{5}{4}\right)^2 = \frac{25}{16} \]

### Step 5: Substitute this value back into the equation
Now, substituting back into the equation:
\[ \frac{1}{E_1^2} + \frac{16}{25} = 1 \]

### Step 6: Isolate \( \frac{1}{E_1^2} \)
Rearranging the equation gives:
\[ \frac{1}{E_1^2} = 1 - \frac{16}{25} \]

### Step 7: Simplify the right side
Calculating the right side:
\[ 1 - \frac{16}{25} = \frac{25}{25} - \frac{16}{25} = \frac{9}{25} \]

### Step 8: Set up the equation for \( E_1^2 \)
Now we have:
\[ \frac{1}{E_1^2} = \frac{9}{25} \]

### Step 9: Invert to find \( E_1^2 \)
Taking the reciprocal gives:
\[ E_1^2 = \frac{25}{9} \]

### Step 10: Take the square root to find \( E_1 \)
Taking the square root of both sides:
\[ E_1 = \frac{5}{3} \]

### Conclusion
Thus, the eccentricity of the conjugate hyperbola is:
\[ \boxed{\frac{5}{3}} \]