1. Given parabola $P_1$

• Vertex: $V=(3,2)$
• Focus: $F=(4,4)$

For a parabola, the vertex is the midpoint of the focus and the foot of the perpendicular from the vertex to the directrix.

So first find the axis direction: $\overrightarrow{VF}=(4-3,4-2)=(1,2)$ Hence the axis is along the vector $(1,2)$.

The focal length is $|VF|=\sqrt{1^2+2^2}=\sqrt{5}$

2. Find the directrix of $P_1$

The directrix is perpendicular to the axis, so its normal vector is along $(1,2)$.

Let the directrix be $x+2y+c=0$ Since the distance from the vertex to the directrix equals $\sqrt{5}$, $\frac{|3+2\cdot 2+c|}{\sqrt{1^2+2^2}}=\sqrt{5}$ $\frac{|7+c|}{\sqrt{5}}=\sqrt{5}$ $|7+c|=5$ So, $c=-2 \quad \text{or} \quad c=-12$

Now determine which one is correct. The focus lies on the side opposite the directrix from the vertex.

For line $x+2y-2=0$:

• At vertex: $3+4-2=5$
• At focus: $4+8-2=10$ Both are on the same side, so this is not the directrix.

For line $x+2y-12=0$:

• At vertex: $3+4-12=-5$
• At focus: $4+8-12=0$ Actually the focus cannot lie on the directrix, so let's use the geometric method directly.

Since the axis direction from vertex to focus is $(1,2)$, the directrix lies on the opposite side of the vertex at distance $\sqrt{5}$. Moving from $V$ opposite to $(1,2)$ by one unit of axis length gives the point $D_0=(3,2)-(1,2)=(2,0)$ This point lies on the directrix, and the directrix is perpendicular to $(1,2)$.

Thus its equation is $1(x-2)+2(y-0)=0$ $x+2y-2=0$ So the directrix of $P_1$ is $x+2y=2$

3. Reflect this directrix about the line $x+2y=6$

The mirror line is $x+2y=6$ The directrix of $P_1$ is $x+2y=2$

These two lines are parallel. Reflection across $x+2y=6$ sends a parallel line at signed distance $d$ on one side to the same distance on the other side.

If the reflected line is $x+2y=k$, then 6 is the midpoint of 2 and $k$: $\frac{2+k}{2}=6$ $2+k=12$ $k=10$

Therefore, the directrix of $P_2$ is $x+2y=10$

4. Comparison with stored answer

Derived answer: $10$

This matches the stored correct answer.