How do you find the vertex, focus and directrix of #y^2+4y+3x-4=0#?

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Answer 1 · 2016-11-10T14:23:15

Vertex is #(8/3,-2)#
Axis of symmetry is #y+2=0#
Focus of parabola is #(31/12,2)#
and directrix is #x=11/4#

#y^2+4y+3x-4=0#

#hArr3x=-(y^2+4y+4)+8#

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

which is equation of a parabola in vertex form whose general equation is #(x-h)=4a(y-k)^2#, whose focus is #(h+a,k)# and directrix is #x=h-a#.

As #h=8/3# and #k=2# vertex is #(8/3,-2)# and axis of symmetry is #y+2=0#. As #a# is negative parabola is open to the left.

As #4a=-1/3#, #a=-1/12# and

focus of parabola is #(8/3-1/12,2)# or #(31/12,2)#

and directrix is #x=8/3+1/12=33/12=11/4#
graph{(y^2+4y+3x-4)=0 [-11.92, 8.08, -6.96, 3.04]}