Chapter 10 - Parametric Equations and Polar Coordinates - 10.5 Conic Sections - 10.5 Exercises - Page 720: 29

Focus: $( 1,-\frac{11}{6})$ and vertex : $(1,-2)$

As we given that $3x^2-6x-2y=1$ or $3(x^2-2x)=2y$ Thus, $3(x-1)^2-4=2y$ Also, $y=\frac{3}{2}(x-1)^2-2$, which shows an equation of an upward parabola with $(h,k+p)=(1,-2+1/6)=( 1,-\frac{11}{6})$ Hence, Focus: $( 1,-\frac{11}{6})$ and vertex : $(1,-2)$