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

Given vertex of parabola is (5, 4) and directrix is  $3x+y-29=0$
Co-ordinates of B (foot of directrix)
 $\frac{x-5}{3}=\frac{y-4}{1}=-\left(\frac{15+4-29}{10}\right)=1$
   
X = 8, y = 5
So, focus will be at same distance to vertex as directrix.
Then, F = (2, 3) (focus)
Equation of parabola is PF = PM with point P(x, y).
 $x^2+9y^2-6xy+134x-2y-711=0$
Compare with given equation. Then,
 $a+b+c+d+k=9-6+134-2-711=-576$