How do you find the vertex, focus and directrix of $( y - 2)^2 = 4(x - 1)$ ?

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Answer 1 · 2018-07-09T06:45:35

Please see the explanation below.

The standard equation of the parabola is

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

Comparing our equation

$(y-2)^2=4(x-1)$

To the standard equation

The vertex is $V=(a,b)=(1,2)$

The focus is $F=(a+p/2,b)=(1+2/2,2)=(2, 2)$

And the directrix is

$x=a-p/2$

$x=1-2/2=0$

graph{((y-2)^2-4(x-1))=0 [-13.11, 15.36, -4.88, 9.36]}

How do you find the vertex, focus and directrix of ( y - 2)^2 = 4(x - 1)( y - 2)^2 = 4(x - 1)?