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. `(x-3)^(2)+(y-(7)/(2))^(2)=(55)/(4)`
B. `(x-3)^(2)+(y-(7)/(2))^(2)=(25)/(4)`
C. `(x-3)^(2)+(y-(7)/(2))^(2)=(7)/(4)`
D. none of these

Explanation

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

### Step 1: Understand the Given Equation
The equation provided is:

\[ |\sqrt{(x-1)^2 + (y-2)^2} - \sqrt{(x-5)^2 + (y-5)^2}| = 3 \]

This represents a hyperbola, where the points \( (1, 2) \) and \( (5, 5) \) are the foci of the hyperbola.

**Hint**: Recognize that the absolute value indicates the difference in distances from a point \( P(x, y) \) to the two foci.

### Step 2: Identify the Foci
The foci of the hyperbola are given by:

- \( S(1, 2) \)
- \( S'(5, 5) \)

**Hint**: The foci are essential for calculating the distance between them, which is used in determining the parameters of the hyperbola.

### Step 3: Calculate the Distance Between the Foci
Using the distance formula, we calculate the distance \( SS' \):

\[ SS' = \sqrt{(5 - 1)^2 + (5 - 2)^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \]

**Hint**: The distance between the foci is denoted as \( 2c \), where \( c \) is the distance from the center of the hyperbola to each focus.

### Step 4: Relate the Distance to the Hyperbola Parameters
For hyperbolas, the relationship between the distance between the foci \( 2c \) and the constant \( 2a \) from the equation is given by:

\[ 2c = 5 \quad \text{and} \quad 2a = 3 \]

From this, we can find \( c \) and \( a \):

\[ c = \frac{5}{2} \quad \text{and} \quad a = \frac{3}{2} \]

**Hint**: Remember that \( c^2 = a^2 + b^2 \) for hyperbolas, where \( b \) is the semi-minor axis.

### Step 5: Calculate \( b \)
Using the relationship \( c^2 = a^2 + b^2 \):

\[ c^2 = \left(\frac{5}{2}\right)^2 = \frac{25}{4} \]
\[ a^2 = \left(\frac{3}{2}\right)^2 = \frac{9}{4} \]

Now substituting:

\[ \frac{25}{4} = \frac{9}{4} + b^2 \]
\[ b^2 = \frac{25}{4} - \frac{9}{4} = \frac{16}{4} = 4 \]
\[ b = 2 \]

**Hint**: This step is crucial for finding the eccentricity of the conjugate hyperbola.

### Step 6: Find the Eccentricity of the Hyperbola
The eccentricity \( e \) of the hyperbola is given by:

\[ e = \frac{c}{a} = \frac{\frac{5}{2}}{\frac{3}{2}} = \frac{5}{3} \]

**Hint**: The eccentricity \( e \) is a measure of how "stretched" the hyperbola is.

### Step 7: Find the Eccentricity of the Conjugate Hyperbola
The relationship between the eccentricity \( e \) of the hyperbola and the eccentricity \( e' \) of the conjugate hyperbola is:

\[ \frac{1}{e^2} + \frac{1}{e'^2} = 1 \]

Substituting \( e = \frac{5}{3} \):

\[ \frac{1}{\left(\frac{5}{3}\right)^2} + \frac{1}{e'^2} = 1 \]
\[ \frac{9}{25} + \frac{1}{e'^2} = 1 \]
\[ \frac{1}{e'^2} = 1 - \frac{9}{25} = \frac{16}{25} \]
\[ e'^2 = \frac{25}{16} \]
\[ e' = \frac{5}{4} \]

**Hint**: This final calculation gives you the eccentricity of the conjugate hyperbola.

### Final Answer
The eccentricity \( e' \) of the corresponding conjugate hyperbola is:

\[ \boxed{\frac{5}{4}} \]