1) We have that the equation is

, hence y=x^2/20. The standard equation of such an equation is y=

. 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

. 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=

.
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:

we get y=

.
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

. Substituting p=7, we get

.
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).

. Thus a=0.378. Hence the correct expressions is y=-0.378*