How do you find the vertex, focus, and directrix of the parabola #(x+1/2)^2=4(y-3)#?

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Answer 1 · 2017-01-05T09:12:18

The vertex is #V=(-1/2,3)#
The focus is #F=(-1/2,4)#
The directrix is #y=2#

The equation is

#(x+1/2)^2=4(y-3)#

We compare this equation to the standard equation of the parabola

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

#p=2#

The vertex is #V=(a,b)=(-1/2,3)#

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

The directrix is #y=b-p/2=3-1=2#

graph{((x+1/2)^2-4(y-3))(y-2)((x+1/2)^2+(y-3)^2-0.01)((x+1/2)^2+(y-4)^2-0.01)=0 [-10, 10, -5, 5]}