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

The parabola y = 1/4d x^2 has
vertex at (0,0)
focal distance = d
focus at (0,1)
directrix y = 1
Now we have transformed our equation to
(x3) = 1/4*4 (y+4)^2
with
vertex at (3,4)
focal distance = 4
focus at (7,4)
directrix x = 1
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