How do you find the vertex, focus and directrix of #y^2+10y-20x+37=0#?

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Answer 1 · 2017-02-13T09:44:41

The vertex is #V=(3/5, -5)#
The focus is #F=(28/5,-5)#
The directrix is #x=-22/5#

Let's rearrange the equation and complete the squares

#y^2+10y-20x+37=0#

#y^2+10y=20x-37#

#y^2+10y+25=20x-37+25#

#(y+5)^2=20x-12=20(x-12/20)=20(x-3/5)#

We compare this equation to the standard equation of a parabola

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

#p=10#

The vertex is #V=(a,b)=(3/5, -5)#

The focus is #F=(a+p/2, b)=(28/5,-5)#

The directrix is #x=a-p/2=3/5-5=-22/5#

graph{(y^2+10y-20x+37)(y+2200x/5+2000)((x-3/5)^2+(y+5)^2-0.01)((x-28/5)^2+(y+5)^2-0.01)=0 [-10, 10, -5, 5]}