What is the equation of the parabola with a focus at (-1,2) and a directrix of #y=1/2#?

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Answer 1 · 2017-11-12T06:07:30

The equation of the parabola is #(x+1)^2=3(y-5/4)#

Any point #(x,y)# on the parabola is equidistant from the directrix and the focus.

Therefore,

#y-(1/2)=sqrt((x-(-1))^2+(y-(2))^2)#

#y-1/2=sqrt((x+1)^2+(y-2)^2)#

Squaring and developing the #(y-2)^2# term and the LHS

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

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

#(x+1)^2=3y-15/4=3(y-5/4)#

The equation of the parabola is #(x+1)^2=3(y-5/4)#

graph{((x+1)^2-3(y-5/4))(y-1/2)((x+1)^2+(y-2)^2-0.04)=0 [-12.66, 12.65, -6.33, 6.33]}