What is the vertex form of the equation of the parabola with a focus at (-4,7) and a directrix of #y=13#?

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Answer 1 · 2016-12-22T15:50:27

The equation is #=-1/12(x+4)^2+10#

The focus is F#=(-4,7)#

and the directrix is #y=13#

By definition, any point #(x,y)# on the parabola is equidistant ffrom the directrix and the focus.

Therefore,

#y-13=sqrt((x+4)^2+(y-7)^2)#

#(y-13)^2=(x+4)^2+(y-7)^2#

#y^2-26y+169=(x+4)^2+y^2-14y+49#

#12y-120=-(x+4)^2#

#y=-1/12(x+4)^2+10#

The parabola opens downwards

graph{(y+1/12(x+4)^2-10)(y-13)=0 [-35.54, 37.54, -15.14, 21.4]}