How do you find the focus, vertex, and directrix of #y^2+4x-4y-8=0#?

1 Answer
Roella W.
Jan 26, 2016

The focus is at#(1,3)#, the vertex is #(2,3)# and the directrix is at #x = 3#

Explanation:

Reformulate the equation to have one variable on its own on the left hand side. In this case it should be #x# because #y# is the one that is raised to a power.

#y^2 +4x - 4y -8=0#
#4x = -y^2+4y+8#
#x=-1/4(y^2 -4y-8)#

This represents a horizontal parabola, because x is being expressed in terms of y. It is opening to the left, because of the minus sign.

We need to get this into vertex form #x= a(y-b)^2 +c# where #(b,c)# is the vertex.

#x= -1/4((y-2)^2-4 -8)#
#x = -1/4(y-2)^2 +12/4 = -1/4(y-2)^2 +3#

The vertex is #(2,3)#

Rearranging the equation as #-4(x-3) = (y-2)^2# which is in the form #4p(x-h) = (y-k)^2# where #p# is the distance between the vertex and the focus which equals the distance from the vertex to the directrix.

In this case #p=-1# so the focus is at#(1,3)# one unit to the left of the vertex.

The directrix is therefore at #x = 3#

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