Question

Fi

Answer 1

my fab stuff lol focal length and all that crab haha 1 moment i'll help

Answer 2

1st of all we have to find the focal length using the following
\[\sqrt{(y - 9)^2}\] and since we have the coordinate (0,-9) the substitute y = -9
\[\sqrt{(-9 - 9)^2}\] which equals 18 but that's the entire focal length we just need half of it to derive the equation so we have now p = 9

NOTE: if the directrex is above the focus then p would turn negative.
so p would turn negative here P = -9

now we need to find a in the equation a(x - h)^2 + k

a = 1/4(p)
a = -1/36

Nowe to find the vertex we do the following
(x value of the focus , y value of the focus +-9 over 2)
so it would be (0, 0)or (0,-9)

So now we have
-1/36 (x - 0)^2 - 9
or
-1/36 (x - 0)^2 + 0

Sorry about the bas explanation but that's best i can do lol.,