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