How do you find the vertex, focus and directrix of #y^2 + 6y + 8x - 7 = 0#?

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Answer 1 · 2017-02-09T09:43:33

The vertex is #V=(2,-3)#
The focus is #F=(0,-3)#
The directrix is #x=4#

Let's rearrage the equation and complete the squares

#y^2+6y+8x-7=0#

#y^2+6y=-8x+7#

#y^2+6y+9=-8x+7+9#

#(y+3)^2=-8x+16=-8(x-2)#

We compare this equation to

#(y-b)^2=2p(x-a)#

#2p=-8#, #=>#, #p=-4#

The vertex is #V=(a,b)=(2,-3)#

The focus is #F=(a+p/2,b)=(0,-3)#

The equation of the directrix is

#x=a-p/2=2+2=4#

graph{(y^2+6y+8x-7)(y+100x-400)((x)^2+(y+3)^2-0.01)=0 [-10, 10, -5, 5]}