Chapter 12 - Parametric Equations, Polar Coordinates, and Conic Sections - 12.5 Conic Sections - Exercises - Page 636: 28

$$x-2=-\frac{1}{16}y^2 .$$

The focus is at $(-2,0)$ and the vertex is $(2,0)$; since $c$ is the distance between the focus and the vertex, then $c=4$. The parabola is horizontal and opens to the left and has the equation $$x-h=-\frac{1}{4c}(y-k)^2\Longrightarrow x-2=-\frac{1}{16}y^2 .$$