How do you find the vertex, focus, and directrix of the parabola #y^2+6y+8x+25=0#?

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Answer 1 · 2017-02-06T07:35:09

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

Rewrite the equation and complete the squares

#y^2+6y+8x+25=0#

#y^2+6y=-8x-25#

#y^2+6y+9=-8x-25+9#

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

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

We compare this equation to

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

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

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

The focus is #(a+p/2,b)=(-4,-3)#

The directrix is #x=a-p/2#, #=>#, #x=-2+2=0#

graph{(y^2+6y+8x+25)((x+2)^2+(y+3)^2-0.1)((x+4)^2+(y+3)^2-0.1)=0 [-59.95, 13.1, -21.5, 15.05]}