Let P_{1} be a parabola with vertex (3,2) and | vedclass

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(A) The axis of the parabola $P_{1}$ passes through the vertex $(3,2)$ and focus $(4,4)$. The slope of the axis is $m = \frac{4-2}{4-3} = 2$.Since the

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$10$

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(A) The axis of the parabola $P_{1}$ passes through the vertex $(3,2)$ and focus $(4,4)$. The slope of the axis is $m = \frac{4-2}{4-3} = 2$.Since the axis is perpendicular to the directrix,the slope of the directrix is $-\frac{1}{2}$.Thus,the equation of the directrix is of the form $x + 2y = k$.The distance from the vertex $(3,2)$ to the directrix is equal to the distance from the vertex to the focus,which is $a = \sqrt{(4-3)^2 + (4-2)^2} = \sqrt{1^2 + 2^2} = \sqrt{5}$.Using the distance formula from point $(3,2)$ to line $x + 2y - k = 0$:$\frac{|3 + 2(2) - k|}{\sqrt{1^2 + 2^2}} = \sqrt{5} \implies |7 - k| = 5$.This gives $7 - k = 5 \implies k = 2$ or $7 - k = -5 \implies k = 12$.Since the focus $(4,4)$ satisfies $4 + 2(4) = 12$,the line $x + 2y = 12$ passes through the focus and cannot be the directrix. Thus,the directrix of $P_{1}$ is $x + 2y = 2$.Let the line of reflection be $L: x + 2y = 6$. The mirror image of the line $x + 2y = 2$ with respect to $x + 2y = 6$ is found by noting that the lines are parallel.If the line $x + 2y = c$ is the image of $x + 2y = 2$ across $x + 2y = 6$,then $6$ is the arithmetic mean of $2$ and $c$:$\frac{2 + c}{2} = 6 \implies 2 + c = 12 \implies c = 10$.Therefore,the directrix of $P_{2}$ is $x + 2y = 10$.

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

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