The vertex of the parabola 2(x−1)2+(y−2)2=(x+y+3) is

2(x−1)2+(y−2)2=(x+y+3)2⇒(x−1)2+(y−2)2=|x+y+3|2 So, focus is S(1,2) and directrix is x+y+3=0 .  Axis of the parabola is x−y+1=0 Solving directrix and axis, we get foot of perpendicular of  directrix on axis as A(−2,−1) .  Therefore, vertex is mid-point of AS which is −12,12 .