How do you find the vertex, focus and directrix of #x^2 - 8x - 16y + 16 = 0#?

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Answer 1 · 2016-12-13T11:18:07

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

Let's rewrite the equation

#x^2-8x+16=16y#

#16y=(x-4)^2#

#y=1/16(x-4)^2#

#(x-4)^2=16y#

This is the equation of a parabola.

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

Where #a=4#, #b=0# and #p=8#

The vertex is #V=(a,b)=(4,0)#

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

And the directrix is #y=b-p/2=0-8/2=-4#

graph{(16y-(x-4)^2)(y+4)=0 [-13.53, 22.5, -4.68, 13.34]}