How do you find the vertex, focus and directrix of #x^2 = 28y#?

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Answer 1 · 2017-01-31T14:02:40

The vertex is #V=(0,0)#
The focus is #F=(0,7)#
The directrix is #y=-7#

Comparing this equation #x^2=28y#

to #x^2=2py#

The vertex is #V=(0,0)#

#2p=28#, #=>#, #p=14#

The focus is #F=(0,p/2)=(0,7)#

The directrix is #y=-p/2#, #=>#, #y=-7#

graph{(28y-x^2)(y+7)((x)^2+(y-7)^2-0.03)=0 [-32.47, 32.5, -16.24, 16.22]}