How do you find vertices, foci, directrix, axis of symmetry of #Y+2 =1/10(x-6)^2#?

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Answer 1 · 2016-11-29T07:22:34

The vertex is #(6,-2)#
The focus is #=(6,1/2)#
The directrix is #y=-9/2#
The axis of symmetry is #x=6#

We rewrite the equation as

#(x-6)^2=10(y+2)#

We compare this to the equation

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

The vertex is #(a,b)=(6,-2)#

#2p=10#, #=>#, #p=5#

The focus is #(a,b+p/2)=(6,-2+5/2)=(6,1/2)#

The directrix is #y=b-p/2=-2-5/2=-9/2#

The axis of symmetry is #x=6#
graph{(x-6)^2=10(y+2) [-36.54, 36.54, -18.27, 18.28]}