How do you find the vertex, focus and directrix of # (x - 1)^2 = 8y - 16#?

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Answer 1 · 2016-12-14T09:02:49

The vertex is #=(1,2)#
The focus is #=(1,4)#
The directrix is #y=0#

Let rewrite the equation as

#(x-1)^2=8(y-2)#

And we compare this equation to the parabola

#(x-a)^2=2p(y-b)#

#2p=8#, #=>#, #p=4#

The vertex is #(a,b)=(1,2)#

The focus is #(a,b+p/2)=(1,4)#

The directrix is #y=b-p/2=2-4/2=0#

graph{(8(y-2)-(x-1)^2)(y)=0 [-14.51, 13.96, -1.41, 12.83]}