Vertex Focus Directrix Calculus Questions

Post ID: 1091695

Overview

Find the vertex, focus, directrix, and focal width of the parabola.

x2 = 12y (1 point)

Vertex: (0, 0); Focus: (3, 0); Directrix: x = 3; Focal width: 3

Vertex: (0, 0); Focus: (0, 3); Directrix: y = -3; Focal width: 12

Vertex: (0, 0); Focus: (3, 0); Directrix: y = 3; Focal width: 48

Vertex: (0, 0); Focus: (0, -3); Directrix: x = -3; Focal width: 48

Find the vertex, focus, directrix, and focal width of the parabola.

x = 3y2 (2 points)

Vertex: (0, 0); Focus: one divided by twelve comma zero ; Directrix: x = ; Focal width: 12

Vertex: (0, 0); Focus: the point one twelfth comma zero ; Directrix: x = negative one twelfth ; Focal width: 0.33

Vertex: (0, 0); Focus: zero comma one divided by sixteen ; Directrix: x = negative one divided by sixteen ; Focal width: 0.33

Vertex: (0, 0); Focus: one divided by sixteen comma zero ; Directrix: y = nnegative one divided by sixteen ; Focal width: 12

Find the standard form of the equation of the parabola with a vertex at the origin and a focus at (0, -7). (1 point)

y = negative one divided by seven x2

y2 = -7x

y = negative one divided by twenty eight x2

y2 = -28x

Find the standard form of the equation of the parabola with a focus at (0, 8) and a directrix at y = -8. (2 points)

y = one divided by thirty two x2

y2 = 8x

y2 = 32x

y = one divided by eight x2

Find the standard form of the equation of the parabola with a focus at (7, 0) and a directrix at x = -7. (2 points)

y = one divided by twenty eight x2

x = one divided by twenty eight y2

-28y = x2

y2 = 14x

A building has an entry the shape of a parabolic arch 96 ft high and 18 ft wide at the base, as shown below.