How do you find the vertex, focus, and directrix of the following parabola and graph the equation #y^2-8y+8x+16=0#?

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Answer 1 · 2017-05-04T05:20:10

Vertex is #(0,4)#, focus is #(-2,4)# and directrix is #x=2#

#y^2-8y+8x+16=0#

#hArr8x=-y^2+8y-16#

or #8x=-(y^2-8y+16)#

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

or #x=-1/8(y-4)^2#

Hence vertex is #(0,4)# and axis of symmetry is #y=4#

As coefficient of #(y-4)^2# is negative, parabola opens on left-hand side and hence focus will be to the left of vertex and directrix on the right.

As this coefficient is #1/8#, equating it to #1/(4p)#, we get #p=2# and focus is #(-2,4)# and directrix is #x=2#

graph{(y^2-8y+8x+16)(x-2)(y-4)((x+2)^2+(y-4)^2-0.04)=0 [-20.25, 19.75, -6.36, 13.64]}