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. `5//3`
B. `4//3`
C. `5//4`
D. `3//sqrt7`

Explanation

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

### Step-by-Step Solution:

1. **Understanding the Given Equation**:
The given equation is:
\[ |\sqrt{(x-1)^2 + (y-2)^2} - \sqrt{(x-5)^2 + (y-5)^2}| = 3 \]
This represents a hyperbola where the difference of distances from two fixed points (foci) is constant.

2. **Identifying the Foci**:
The foci of the hyperbola can be identified as:
- Focus 1 (S): \( (1, 2) \)
- Focus 2 (S'): \( (5, 5) \)

3. **Finding the Distance Between the Foci**:
The distance \( SS' \) between the two foci can be calculated using the distance formula:
\[ SS' = \sqrt{(5 - 1)^2 + (5 - 2)^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \]

4. **Relating the Distance to the Hyperbola Parameters**:
For a hyperbola, the relationship between the distance between the foci \( 2c \) and the constant \( 2a \) (which is the given constant 3 in this case) is:
\[ 2c = SS' = 5 \quad \text{and} \quad 2a = 3 \]
From this, we can find \( c \) and \( a \):
\[ c = \frac{5}{2}, \quad a = \frac{3}{2} \]

5. **Finding the Value of \( b \)**:
The relationship between \( a \), \( b \), and \( c \) for a hyperbola is given by:
\[ c^2 = a^2 + b^2 \]
Plugging in the values:
\[ \left(\frac{5}{2}\right)^2 = \left(\frac{3}{2}\right)^2 + b^2 \]
\[ \frac{25}{4} = \frac{9}{4} + b^2 \]
\[ b^2 = \frac{25}{4} - \frac{9}{4} = \frac{16}{4} = 4 \quad \Rightarrow \quad b = 2 \]

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

7. **Finding the Eccentricity of the Conjugate Hyperbola**:
The relationship between the eccentricity \( e \) of a hyperbola and the eccentricity \( e' \) of its conjugate hyperbola is given by:
\[ \frac{1}{e^2} + \frac{1}{(e')^2} = 1 \]
Substituting \( e = \frac{5}{3} \):
\[ \frac{1}{\left(\frac{5}{3}\right)^2} + \frac{1}{(e')^2} = 1 \]
\[ \frac{9}{25} + \frac{1}{(e')^2} = 1 \]
\[ \frac{1}{(e')^2} = 1 - \frac{9}{25} = \frac{16}{25} \]
\[ (e')^2 = \frac{25}{16} \quad \Rightarrow \quad e' = \frac{5}{4} \]

### Final Answer:
The eccentricity \( e' \) of the corresponding conjugate hyperbola is:
\[ \boxed{\frac{5}{4}} \]