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]}

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