Let `P(x, y)` is a variable point such that `|sqrt((x-1)^2+(y-2)^2)-sqrt((x-5)^2+(y-5)^2)|=3` , which represents hyperbola. The eccentricity e' of the corresponding conjugate hyperbola is (A) `5/3` (B) `4/3` (C) `5/4` (D) `3/sqrt7`

Options

A. `tan^(-1)(4//3)`
B. `tan^(-1)(3//4)`
C. `tan^(-1)(5//3)`
D. `tan^(-1)(3//5)`

Explanation

To solve the problem, we need to analyze the given equation of the hyperbola and find the eccentricity of its conjugate hyperbola.

### Step-by-Step Solution:

1. **Understanding the Given Equation**:
We have the equation:
\[ | \sqrt{(x-1)^2 + (y-2)^2} - \sqrt{(x-5)^2 + (y-5)^2} | = 3 \]
This represents a hyperbola where the points \( S(1, 2) \) and \( S'(5, 5) \) are the foci.

2. **Identifying the Foci**:
Let \( S = (1, 2) \) and \( S' = (5, 5) \). The distance between the foci \( S \) and \( S' \) can be calculated using the distance formula:
\[ d = \sqrt{(5 - 1)^2 + (5 - 2)^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \]

3. **Relating to Hyperbola Properties**:
For a hyperbola, the difference of distances from any point \( P(x, y) \) to the foci is a constant. This constant is given as 3, which can be expressed as:
\[ 2a = 3 \implies a = \frac{3}{2} \]

4. **Finding the Value of \( c \)**:
The distance between the foci \( 2c \) is given by:
\[ 2c = 5 \implies c = \frac{5}{2} \]

5. **Finding \( b \)**:
We know the relationship in hyperbolas:
\[ c^2 = a^2 + b^2 \]
Substituting the values we have:
\[ \left(\frac{5}{2}\right)^2 = \left(\frac{3}{2}\right)^2 + b^2 \]
This simplifies to:
\[ \frac{25}{4} = \frac{9}{4} + b^2 \implies b^2 = \frac{25}{4} - \frac{9}{4} = \frac{16}{4} = 4 \implies b = 2 \]

6. **Finding the Eccentricity of the Conjugate Hyperbola**:
The eccentricity \( e' \) of the conjugate hyperbola is given by:
\[ e' = \frac{c}{a} \]
Substituting the values of \( c \) and \( a \):
\[ e' = \frac{\frac{5}{2}}{\frac{3}{2}} = \frac{5}{3} \]

### Conclusion:
The eccentricity \( e' \) of the corresponding conjugate hyperbola is:
\[ \boxed{\frac{5}{3}} \]