Chapter 12 - Parametric Equations, Polar Coordinates, and Conic Sections - 12.5 Conic Sections - Exercises - Page 636: 35

Parabola with: - focus $F = (4, \frac{1}{16}),$ - directrix $y = −\frac{1}{16}$, - vertex $ (4, 0).$

The equation $y=4(x-4)^2$ is a parabola and by comparing it with the standard equation $y-k=\frac{1}{4c}(x-h)^2$, we get $4=\frac{1}{4c}$ $c=\frac{1}{16}$ Thus, we have: - Focus $F = (4, \frac{1}{16}),$ - Directrix $y = −\frac{1}{16}$, - Vertex at the origin $ (4, 0).$