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(D) The vertex $A$ is $(5,4)$ and the directrix is $3x+y-29=0$.Let $B$ be the foot of the perpendicular from the vertex to the directrix. The line pas

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$-576$

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(D) The vertex $A$ is $(5,4)$ and the directrix is $3x+y-29=0$.Let $B$ be the foot of the perpendicular from the vertex to the directrix. The line passing through $A$ and perpendicular to the directrix has the equation $\frac{x-5}{3} = \frac{y-4}{1} = k$.Since $B$ lies on the directrix $3x+y-29=0$,we have $3(5+3k) + (4+k) - 29 = 0$,which gives $15+9k+4+k-29=0$,so $10k-10=0$,implying $k=1$.Thus,the coordinates of $B$ are $(5+3(1), 4+1) = (8,5)$.Since the vertex $A$ is the midpoint of the segment $SB$,where $S$ is the focus $(x_s, y_s)$,we have $\frac{x_s+8}{2} = 5$ and $\frac{y_s+5}{2} = 4$,which gives $S = (2,3)$.The definition of a parabola is the locus of points $P(x,y)$ such that $PS = PM$,where $PM$ is the perpendicular distance to the directrix.$PS^2 = (x-2)^2 + (y-3)^2 = x^2-4x+4+y^2-6y+9 = x^2+y^2-4x-6y+13$.$PM^2 = \frac{(3x+y-29)^2}{3^2+1^2} = \frac{9x^2+y^2+841+6xy-174x-58y}{10}$.Equating $10(x^2+y^2-4x-6y+13) = 9x^2+y^2+6xy-174x-58y+841$,we get $x^2+9y^2-6xy+134x-2y-711=0$.Comparing with $x^2+ay^2+bxy+cx+dy+k=0$,we have $a=9, b=-6, c=134, d=-2, k=-711$.Therefore,$a+b+c+d+k = 9-6+134-2-711 = -576$.

If the equation of the parabola,whose vertex is at $(5,4)$ and the directrix is $3x+y-29=0$,is $x^{2}+ay^{2}+bxy+cx+dy+k=0$,then $a+b+c+d+k$ is equal to

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