Let `P_(1)` be a parabola with vertex (3,2) and focus (4,4) and `P_(2)` be its mirror image with respect to the line `x+2y=6` . Then the directrix of `P_(2)` is `x+2y `= ________.

Answer

To solve the problem, we need to find the directrix of the parabola \( P_2 \), which is the mirror image of the parabola \( P_1 \) with respect to the line \( x + 2y = 6 \). Let's break down the solution step by step.

### Step 1: Understand the properties of the parabola \( P_1 \)

The parabola \( P_1 \) has:
- Vertex: \( (3, 2) \)
- Focus: \( (4, 4) \)

The directrix of a parabola is a line that is equidistant from the vertex as the focus but in the opposite direction.

### Step 2: Find the equation of the directrix of \( P_1 \)

The distance from the vertex to the focus can be calculated as follows:
- The distance between the vertex \( (3, 2) \) and the focus \( (4, 4) \) is:
\[ d = \sqrt{(4 - 3)^2 + (4 - 2)^2} = \sqrt{1 + 4} = \sqrt{5} \]

Since the vertex is at \( (3, 2) \) and the focus is at \( (4, 4) \), the directrix will be located at a distance of \( \sqrt{5} \) in the opposite direction from the vertex.

### Step 3: Determine the direction from the vertex to the focus

The direction vector from the vertex to the focus is:
\[ (4 - 3, 4 - 2) = (1, 2) \]

To find the unit vector in this direction, we normalize it:
\[ \text{Unit vector} = \left( \frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}} \right) \]

### Step 4: Find the coordinates of the directrix

To find the coordinates of the directrix, we move \( \sqrt{5} \) units in the opposite direction of the unit vector:
\[ \text{Directrix point} = (3, 2) - \sqrt{5} \left( \frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}} \right) = (3, 2) - (1, 2) = (2, 0) \]

### Step 5: Write the equation of the directrix

The directrix can be expressed in the form \( x + 2y = k \). To find \( k \), we can substitute the point \( (2, 0) \):
\[ 2 + 2(0) = k \implies k = 2 \]

Thus, the equation of the directrix \( D_1 \) of the parabola \( P_1 \) is:
\[ x + 2y = 2 \]

### Step 6: Find the directrix of the mirror image parabola \( P_2 \)

The line of reflection is given by \( x + 2y = 6 \). The distance between the directrix \( D_1 \) and the line of reflection is:
\[ \text{Distance} = \frac{|2 - 6|}{\sqrt{1^2 + 2^2}} = \frac{4}{\sqrt{5}} \]

Since \( P_2 \) is the mirror image, we need to move this distance \( \frac{4}{\sqrt{5}} \) in the opposite direction from the line of reflection.

### Step 7: Calculate the new directrix \( D_2 \)

The new directrix \( D_2 \) will be:
\[ x + 2y = 6 + \frac{4}{\sqrt{5}} \cdot \sqrt{5} = 6 + 4 = 10 \]

Thus, the equation of the directrix \( D_2 \) of the parabola \( P_2 \) is:
\[ x + 2y = 10 \]

### Final Answer

The directrix of \( P_2 \) is:
\[ \boxed{10} \]