Chapter 10 - Section 10.4 - The Parabola; Identifying Conic Sections - Exercise Set - Page 799: 13

$(0,6)$

Standard form of a horizontal parabola is $x=a(y-k)^2+h$ Here, vertex:$(h,k)$ and $4a$ is latus rectum. Since we are given $x=2(y-6)^2$ It can be written as: $x=2(y-6)^2+0$ Compare the given equation with the standard form. Thus, vertex: $(0,6)$