precalc
posted by 95 .
what is the directrix and focus of the equation 1/16(y+4)^2=x3

precalc 
drwls
Perform a substitution
x' = x3
y' = y+4
and you are left with the equatiom
y'^2 = 16x'
This is the equation of a parabola with axis along the line y = 4 and vertex at (x',y') = (0,0)
which is (x,y) = (3,4)
The factor 16 tells you where the focus and directrix are.
16 = 4p,
where p is the distance from the vertex to the focus and directrix.
That puts the focus at (x,y) = (7,4)
The directrix is the vertical line x = 1
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