3.) Write an equation of the parabola with

a.) focus (0, –2) and directrix y = 2

The directrix is above the focus so this parabola opens downward

The vertex is   halfway between the focus and the directrix....so it must be (0, 0)

"p" is the distance between the vertex and the focus [ or the vertex and the directrix ] =  2

So.....we have this form

4p ( y - k) = - ( x - h)^2         where (h, k) is the vertex and p = 2

So we have

4(2) ( y - 0) = - ( x - 0)^2          simplify

8y = -x^2         divide both sides by 8

y = -(1/8)x^2

b.) focus (3, 5) and directrix y = 7

Again....the directrix is above the focus....so this turns downward

The vertex is hafway between the focus and directrix = (3, 6)

p =   distance between focus and vertex = 1

So we have

4p (y - k) =  -(x - h)^2             (h, k) = (3, 6)   and p = 1

4(1) (y - 6) = - (x - 3)^2

4 ( y - 6) = - ( x - 3)^2

y - 6 =  - ( 1/4)(x - 3)^2

y =   - (1/4) ( x - 3)^2 + 6

c.) focus (–2, 1) and directrix y = 13

turns downward

vertex =   ( - 2 ,  (13 + 1) / 2 )   =  ( -2, 14/2)   = ( -2, 7)

p =   6

4p ( y - 7) = - (x - - 2 )^2

4(6) ( y - 7) = - ( x + 2)^2

24 ( y - 7) = - ( x + 2)^2

y - 7=  - (1/24) ( x + 2)^2

y = - (1/24)( x + 2)^2 + 7