Given vertex is (5,4) and directrix 3x+y−29=0Let foot of perpendicular of (5,4) on directrix is (x1,y1)3x1−5=1y1−4=10−(−10)∴(x1,y1)≡(8,5)So, focus of parabola will be S=(2,3)Let P(x,y) be any point on parabola, then(x−2)2+(y−3)2=10(3x+y−29)2⇒10(x2+y2−4x−6y+13)=9x2+y2+841+6xy−58y−174x⇒x2+9y2−6xy+134x−2y−711=0and given parabolax2+ay2+bxy+cx+dy+k=0∴a=9,b=−6,c=134,d=−2,k=−711∴a+b+c+d+k=−576