Statement-I `(5)/(3) and (5)/(4)` are the eccentricities of two conjugate hyperbolas.
Statement-II If `e_1 and e_2` are the eccentricities of two conjugate hyperbolas, then `e_1e_2gt1`.

Options

A. Statement-I is true, Statement-II is also true, Statement-II is the correct explanation of Statement-I.
B. Statement-I is true, Statement-II is also true, Statement-II is not the correct explanation of Statement-I.
C. Statement-I is true, Statement-II is false.
D. Statement-I is false, Statement-II is true

Answer

To solve the problem, we need to analyze the two statements provided regarding the eccentricities of conjugate hyperbolas.

### Step-by-Step Solution:

**Step 1: Analyze Statement I**
- Statement I claims that \(\frac{5}{3}\) and \(\frac{5}{4}\) are the eccentricities of two conjugate hyperbolas.
- Let \(e_1 = \frac{5}{3}\) and we need to find \(e_2\).

**Step 2: Use the Property of Eccentricities of Conjugate Hyperbolas**
- The property states that for two conjugate hyperbolas, the relationship between their eccentricities is given by:
\[ \frac{1}{e_1^2} + \frac{1}{e_2^2} = 1 \]

**Step 3: Substitute \(e_1\) into the Equation**
- Substitute \(e_1 = \frac{5}{3}\):
\[ \frac{1}{\left(\frac{5}{3}\right)^2} + \frac{1}{e_2^2} = 1 \]
- Calculate \(\left(\frac{5}{3}\right)^2\):
\[ \left(\frac{5}{3}\right)^2 = \frac{25}{9} \]
- Therefore, we have:
\[ \frac{1}{\frac{25}{9}} + \frac{1}{e_2^2} = 1 \]
- This simplifies to:
\[ \frac{9}{25} + \frac{1}{e_2^2} = 1 \]

**Step 4: Solve for \(e_2^2\)**
- Rearranging gives:
\[ \frac{1}{e_2^2} = 1 - \frac{9}{25} \]
- Convert \(1\) into a fraction:
\[ 1 = \frac{25}{25} \]
- Thus:
\[ \frac{1}{e_2^2} = \frac{25}{25} - \frac{9}{25} = \frac{16}{25} \]

**Step 5: Calculate \(e_2\)**
- Taking the reciprocal gives:
\[ e_2^2 = \frac{25}{16} \]
- Taking the square root:
\[ e_2 = \frac{5}{4} \]

**Conclusion for Statement I:**
- Since we have found \(e_2 = \frac{5}{4}\), Statement I is **True**.

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**Step 6: Analyze Statement II**
- Statement II claims that if \(e_1\) and \(e_2\) are the eccentricities of two conjugate hyperbolas, then \(e_1 e_2 > 1\).
- We know that both \(e_1\) and \(e_2\) must be greater than 1 for hyperbolas.

**Step 7: Verify the Product of Eccentricities**
- Calculate the product:
\[ e_1 e_2 = \left(\frac{5}{3}\right) \left(\frac{5}{4}\right) = \frac{25}{12} \]
- Since \(\frac{25}{12} > 1\), Statement II is also **True**.

### Final Conclusion:
- Both statements are true, but Statement II does not serve as a correct explanation for Statement I.