What is the equation of a parabola with a vertex at #(0,-1/4)#and a directrix of #(0,1/4)#?

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Answer 1 · 2018-05-21T06:59:08

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

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

The focus is #F=(0,-3/4)#

Therefore,

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

#x^2+y^2+3/2y+9/16=y^2-1/2y+1/16#

#x^2=-2y-1/2#

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

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

graph{(y+x^2/2+1/4)(y-1/4)=0 [-3.462, 3.466, -1.732, 1.732]}