The locus of the foot of perpendicular from my focus of a hyperbola upon any tangent to the hyperbola is the auxiliary circle of the hyperbola. Consider the foci of a hyperbola as `(-3, -2)` and (5,6) and the foot of perpendicular from the focus (5, 6) upon a tangent to the hyperbola as (2, 5).
The conjugate axis of the hyperbola is

Options

A. `4sqrt(11)`
B. `2sqrt(11)`
C. `4sqrt(22)`
D. `2sqrt(22)`

Explanation

To find the conjugate axis of the hyperbola given the foci and the foot of the perpendicular from one of the foci to a tangent, we will follow these steps:

### Step 1: Find the Center of the Hyperbola
The center of the hyperbola can be found by taking the midpoint of the line segment joining the foci.

Given foci are:
- Focus 1: \( F_1 = (-3, -2) \)
- Focus 2: \( F_2 = (5, 6) \)

The center \( C \) is given by:
\[ C = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) = \left( \frac{-3 + 5}{2}, \frac{-2 + 6}{2} \right) = \left( \frac{2}{2}, \frac{4}{2} \right) = (1, 2) \]

### Step 2: Determine the Radius of the Auxiliary Circle
The foot of the perpendicular from the focus \( (5, 6) \) to the tangent is given as \( (2, 5) \). This point lies on the auxiliary circle.

The radius \( a \) of the auxiliary circle can be calculated using the distance formula:
\[ a = \sqrt{(x_C - x_T)^2 + (y_C - y_T)^2} \]
Where \( (x_C, y_C) \) is the center \( (1, 2) \) and \( (x_T, y_T) \) is the foot of the perpendicular \( (2, 5) \).

Calculating the radius:
\[ a = \sqrt{(1 - 2)^2 + (2 - 5)^2} = \sqrt{(-1)^2 + (-3)^2} = \sqrt{1 + 9} = \sqrt{10} \]

### Step 3: Calculate the Distance Between the Foci
The distance \( d \) between the two foci is given by:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{(5 - (-3))^2 + (6 - (-2))^2} = \sqrt{(5 + 3)^2 + (6 + 2)^2} = \sqrt{8^2 + 8^2} = \sqrt{64 + 64} = \sqrt{128} = 8\sqrt{2} \]

### Step 4: Relate the Distance Between Foci to the Eccentricity
The distance between the foci is also given by the formula:
\[ d = 2ae \]
Where \( e \) is the eccentricity and \( a \) is the semi-major axis length.

From the previous step, we have:
\[ 8\sqrt{2} = 2a \cdot e \]

### Step 5: Find the Eccentricity
We can express \( e \) in terms of \( a \):
\[ e = \frac{8\sqrt{2}}{2a} = \frac{4\sqrt{2}}{a} \]

### Step 6: Find the Value of \( b \)
Using the relationship \( e^2 = 1 + \frac{b^2}{a^2} \):
\[ \left( \frac{4\sqrt{2}}{a} \right)^2 = 1 + \frac{b^2}{a^2} \]
\[ \frac{32}{a^2} = 1 + \frac{b^2}{a^2} \]
Multiplying through by \( a^2 \):
\[ 32 = a^2 + b^2 \]
We know \( a^2 = 10 \), so:
\[ 32 = 10 + b^2 \implies b^2 = 22 \]

### Step 7: Calculate the Conjugate Axis
The conjugate axis is given by:
\[ 2b = 2\sqrt{22} \]

### Final Answer
Thus, the conjugate axis of the hyperbola is:
\[ \boxed{2\sqrt{22}} \]