8.01

1. find the vertex, focus, directrix, and focal width of the parabola.
x2 = 20y

a) vertex: (0, 0); focus: (0, 5); directrix: y = -5; focal width: 20
b) vertex: (0, 0); focus: (5, 0); directrix: x = 5; focal width: 5
c) vertex: (0, 0); focus: (5, 0); directrix: y = 5; focal width: 80
d) vertex: (0, 0); focus: (0, -5); directrix: x = -5; focal width: 80

2. find the vertex, focus, directrix, and focal width of the parabola.
x = 3y2

a) vertex: (0, 0); focus: one divided by twelve comma zero ; directrix: x = one divided by twelve ; focal width: 12
b) vertex: (0, 0); focus: the point one twelfth comma zero ; directrix: x = negative one twelfth ; focal width: 0.33
c) vertex: (0, 0); focus: zero comma one divided by sixteen ; directrix: x = negative one divided by sixteen ; focal width: 0.33
d) vertex: (0, 0); focus: one divided by sixteen comma zero ; directrix: y = nnegative one divided by sixteen ; focal width: 12

3. find the standard form of the equation of the parabola with a vertex at the origin and a focus at (0, 9). (1 point)

a) y = one divided by thirty six x2
b) y = one divided by nine x2
c) y2 = 9x
d) y2 = 36x

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

a) y = one divided by thirty two x2
b) y2 = 8x
c) y2 = 32x
d) y = one divided by eight x2

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

a) y = one divided by twenty eight x2
b) x = one divided by twenty eight y2
c) -28y = x2
d) y2 = 14x

6. a building has an entry the shape of a parabolic arch 74 ft high and 28 ft wide at the base, as shown below.
a parabola opening down with vertex at the origin is graphed on the coordinate plane. the height of the parabola from top to bottom is seventy four feet and its width from left to right is twenty eight feet.
find an equation for the parabola if the vertex is put at the origin of the coordinate system.