How do you find the vertex, focus and directrix of #4x-y^2-2y-33=0#?

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Answer 1 · 2017-12-23T09:00:43

Vertex is #(8,-1)#; focus is #(9,-1)# and directrix is #x=7#

Let us rewrite #4x-y^2-2y-33=0# as

#4x=y^2+2y+33#

i.e. #4x=(y+1)^2+32#

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

Hence, vertex is #(8,-1)# and axis of symmetry is #y+1=0#

If the equation is of the type #(y-k)^2=4p(x-h)#,

the focus is #(h+p,k)# and the directrix is #x=h-p#

Here it is of the form #(y+1)^2=4(x-8)#

Hence, #k=-1#, #h=8# and #p=1#

Hence focus is #(9,-1)# and directrix is #x=7#

Parabola appears as

graph{(4x-y^2-2y-33)((x-8)^2+(y+1)^2-0.01)((x-9)^2+(y+1)^2-0.01)(x-7)=0 [2.35, 12.35, -3.46, 1.54]}