How do you find the vertex, focus and directrix of #x =-1/2(y-2)^2-4#?

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Answer 1 · 2018-04-09T12:57:02

Vertex is at #(-4,2)#, focus is at #(-4.5,2) # and
directrix is # x=-3.5#

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

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

Comparing with the equation of horizontal

parabola opening left is #(y-k)^2 = -4p(x-h)#

We get , #h=-4 ,k=2, p=1/2 :.# vertex is at

#(h,k) or (-4,2)# , focus is at #(-4-1/2),2or (-4.5,2) #

and directrix is # x= (-4+1 /2) or x = -3.5#

graph{x= -1/2(y-2)^2-4 [-10, 10, -5, 5]}