Derive the equation of the parabola with a focus at (6, 2) and a directrix of y = 1.
f(x) = −1/2 times (x − 6)^2 + 3/2
f(x) = 1/2 times (x − 6)^2 + 3/2
f(x) = −1/2 times (x + 3/2)^2 + 6
f(x) = 1/2 times (x + 3/2)^2 + 6

(x-h)^2=4p(y-k)
we know it is this one because the directix is y=something and not x=something

(h,k) is the vertex
p is the distance from the vertex to the focus
it is 1/2 the distance from the vertex to the directix

if p is positive, the focus is above the vertex
if p is negative, the focus is below the vertex

we have
focus is (6,2) and directix is y=1
distance from (6,2) to y=1 is the distance from 2 to 1 which is 1
1/2=p
since 1<2, we see that the focus is above the vertex (when the focus is greater than the directix, then the graph opens to the right or up)

p=1/2
1/2 below (6,2) is (6,1.5)
vertex is (6,1.5)
p=1/2

(x-6)^2=4(0.5)(y-1.5)
(x-6)^2=2(y-1.5)
(x-6)^2=2y-3
2y=(x-6)^2+3
divide both sides by 2
y=1/2 times (x-6)^2+3/2
f(x)=1/2 times (x-6)^2+3/2

2nd option is answer

Idk idk idk idk idk idk

Set square i think so