Chapter 12 - Parametric Equations, Polar Coordinates, and Conic Sections - 12.5 Conic Sections - Exercises - Page 636: 34

The graph is a parabola with: - Focus $F = (0, \frac{1}{16}),$ - Directrix $y = −\frac{1}{16}$, - Vertex $ (0, 0).$

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