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

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Answer 1 · 2017-03-16T14:25:32

The vertex is #V=(0,2)#
The focus is #F=(1/8,2)#
The directrix is #x=-1/8#

Let's rearrange the equation by completing the squares

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

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

#2(y^2-4y)=x-8#

#2(y^2-4y+4)=x-8+8#

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

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

This is the equation of a parabola.

We compare this equation to

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

#p=1/4#

The vertex is #V=(a,b)=(0,2)#

The focus is #F=(a+p/2,b)=(1/8,2)#

The directrix is #x=-1/8#

graph{((y-2)^2-x/2)(y-1000(x+1/8))((x-1/8)^2+(y-2)^2-0.001)=0 [-5.222, 5.874, -0.687, 4.86]}