PLEASE HELP8.011. 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: 802. 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: 123. 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 = 36x4. 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 x25. 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 = 14x6. 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.

1) We have that the equation is [tex]x^2=20y[/tex] , hence y=x^2/20. The standard equation of such an equation is y=[tex] \frac{1}{4p} x^2[/tex]. Hence, p=5 in this case. The focus is at (0,5) and the directrix is at y=-5 (a tip is that the directrix is always "opposite" the focus point of a parabola; if the directrix is at x=-7 for example, the focus is at (7,0)).
2) Similarly, we have that the equation is [tex]x=3y^2 \\ \frac{1}{4p} =3[/tex]. Thus, p=1/12. In this case, the parabola opens along the x-axis and the focus is at (1/12, 0). Also, the directrix is at x=-1/12. Hence the correct answer is B.
3) We are given that the parabola has a p of 9. Also, the focus lies along the y-axis, hence the parabola is opening along the y-axis. Finally, the focus is on the positive half, so the parabola is opening upwards. The equation for this case is y=[tex] y=\frac{1}{4p} x^2= \frac{1}{36 } x^2[/tex].
4) Similarly as above. The directrix is superfluous, we only need the p-value. THe same comments about the parabola apply and if we substitute p=8 in the formula: [tex]y= \frac{1}{4p} x^2 [/tex] we get y=[tex] \frac{1}{32} x^2[/tex].
5) This is somewhat different, even though we do not need the directrix again. The focus lies on the x-axis, thus the parabola opens in this direction. The focus lies on the positive part of the axis, thus the parabola opens to the right. We also are given p=7. Hence, the equation we need is of the form[tex]x= \frac{1}{4p} y^2[/tex]. Substituting p=7, we get [tex]x= \frac{1}{28} y^2[/tex].
6) The equation of a prabola with a vertex at (0,0) is of the form y=-ax^2. The minus sign is needed since the parabola is downwards. Since we are given anothe point, we can calculate a. We have to take y=-74 and x=14 feet (since left to right is 28, we need to take half). [tex]-a= \frac{y}{x^2} = \frac{-74}{14^2} =-0.378[/tex]. Thus a=0.378. Hence the correct expressions is y=-0.378*[tex]x^2[/tex]