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 directrix of the hyperbola corresponding to the focus (5, 6) is

Options

A. `2x+2y-1=0`
B. `2x+2y-11=0`
C. `2x+2y-7=0`
D. `2x+2y-9=0`

Explanation

To find the directrix of the hyperbola corresponding to the focus (5, 6), we will follow these steps:

### Step 1: Find the Center of the Hyperbola
The center of the hyperbola can be found by calculating the midpoint of the two foci, which are given as (-3, -2) and (5, 6).

\[ \text{Center} = \left( \frac{-3 + 5}{2}, \frac{-2 + 6}{2} \right) = \left( \frac{2}{2}, \frac{4}{2} \right) = (1, 2) \]

### Step 2: Verify the Foot of the Perpendicular
We are given the foot of the perpendicular from the focus (5, 6) to the tangent as (2, 5). We need to confirm that this point lies on the auxiliary circle of the hyperbola.

The radius of the auxiliary circle is the distance from the center to one of the foci. We can calculate this distance:

\[ \text{Radius} = \sqrt{(5 - 1)^2 + (6 - 2)^2} = \sqrt{4^2 + 4^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2} \]

### Step 3: Calculate the Radius of the Auxiliary Circle
The radius of the auxiliary circle is given by \( a \), where \( a \) is the semi-major axis of the hyperbola.

Using the coordinates of the center (1, 2) and the foot of the perpendicular (2, 5):

\[ \text{Distance} = \sqrt{(2 - 1)^2 + (5 - 2)^2} = \sqrt{1^2 + 3^2} = \sqrt{1 + 9} = \sqrt{10} \]

### Step 4: Find the Eccentricity
The distance between the two foci is calculated as follows:

\[ \text{Distance between foci} = \sqrt{(-3 - 5)^2 + (-2 - 6)^2} = \sqrt{(-8)^2 + (-8)^2} = \sqrt{64 + 64} = \sqrt{128} = 8\sqrt{2} \]

Let \( 2ae = 8\sqrt{2} \) where \( a = \sqrt{10} \).

\[ 2 \cdot \sqrt{10} \cdot e = 8\sqrt{2} \]
\[ e = \frac{8\sqrt{2}}{2\sqrt{10}} = \frac{4\sqrt{2}}{\sqrt{10}} = \frac{4\sqrt{2}}{\sqrt{2 \cdot 5}} = \frac{4}{\sqrt{5}} = \frac{4\sqrt{5}}{5} \]

### Step 5: Find the Directrix
The directrix corresponding to the focus (5, 6) can be expressed as:

\[ x + y = \lambda \]

The distance from the center (1, 2) to the directrix is given by:

\[ \frac{1 + 2 - \lambda}{\sqrt{2}} = \frac{5}{2\sqrt{2}} \]

Setting the two distances equal gives:

\[ \frac{3 - \lambda}{\sqrt{2}} = \frac{5}{2\sqrt{2}} \]

Cross-multiplying:

\[ 3 - \lambda = \frac{5}{2} \]
\[ \lambda = 3 - \frac{5}{2} = \frac{6}{2} - \frac{5}{2} = \frac{1}{2} \]

Thus, the equation of the directrix is:

\[ x + y = \frac{11}{2} \]

Rearranging gives:

\[ 2x + 2y = 11 \]

### Final Answer
The directrix of the hyperbola corresponding to the focus (5, 6) is:

\[ 2x + 2y - 11 = 0 \]